Quadratic Equation Compiled

70 MATHEMATICS

44.1 Introduction

In Chapter 2, you have studied different types of polynomials. One type was the

quadratic polynomial of the form ax2 + bx + c, a ! 0. When we equate this polynomial

to zero, we get a quadratic equation. Quadratic equations come up when we deal with

many real-life situations. For instance, suppose a

charity trust decides to build a prayer hall having

a carpet area of 300 square metres with its length

one metre more than twice its breadth. What

should be the length and breadth of the hall?

Suppose the breadth of the hall is x metres. Then,

its length should be (2x + 1) metres. We can depict

this information pictorially as shown in Fig. 4.1.

Now, area of the hall = (2x + 1). x m2 = (2x2 + x) m2

So, 2x2 + x = 300 (Given)

Therefore, 2x2 + x – 300 = 0

So, the breadth of the hall should satisfy the equation 2x2 + x – 300 = 0 which is a

quadratic equation.

Many people believe that Babylonians were the first to solve quadratic equations.

For instance, they knew how to find two positive numbers with a given positive sum

and a given positive product, and this problem is equivalent to solving a quadratic

equation of the form x2 – px + q = 0. Greek mathematician Euclid developed a

geometrical approach for finding out lengths which, in our present day terminology,

are solutions of quadratic equations. Solving of quadratic equations, in general form, is

often credited to ancient Indian mathematicians. In fact, Brahmagupta (A.D.598–665)

gave an explicit formula to solve a quadratic equation of the form ax2 + bx = c. Later,

QUADRATIC EQUATIONS

Fig. 4.1

QUADRATIC EQUATIONS 71

Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula,

(as quoted by Bhaskara II) for solving a quadratic equation by the method of completing

the square. An Arab mathematician Al-Khwarizmi (about A.D. 800) also studied

quadratic equations of different types. Abraham bar Hiyya Ha-Nasi, in his book

‘Liber embadorum’ published in Europe in A.D. 1145 gave complete solutions of

different quadratic equations.

In this chapter, you will study quadratic equations, and various ways of finding

their roots. You will also see some applications of quadratic equations in daily life

situations.

4.2 Quadratic Equations

A quadratic equation in the variable x is an equation of the form ax2 + bx + c = 0, where

a, b, c are real numbers, a ! 0. For example, 2x2 + x – 300 = 0 is a quadratic equation.

Similarly, 2x2 – 3x + 1 = 0, 4x – 3x2 + 2 = 0 and 1 – x2 + 300 = 0 are also quadratic

equations.

In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree

2, is a quadratic equation. But when we write the terms of p(x) in descending order of

their degrees, then we get the standard form of the equation. That is, ax2 + bx + c = 0,

a ! 0 is called the standard form of a quadratic equation.

Quadratic equations arise in several situations in the world around us and in

different fields of mathematics. Let us consider a few examples.

Example 1 : Represent the following situations mathematically:

(i) John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and

the product of the number of marbles they now have is 124. We would like to find

out how many marbles they had to start with.

(ii) A cottage industry produces a certain number of toys in a day. The cost of

production of each toy (in rupees) was found to be 55 minus the number of toys

produced in a day. On a particular day, the total cost of production was

Rs 750. We would like to find out the number of toys produced on that day.

Solution :

(i) Let the number of marbles John had be x.

Then the number of marbles Jivanti had = 45 – x (Why?).

The number of marbles left with John, when he lost 5 marbles = x – 5

The number of marbles left with Jivanti, when she lost 5 marbles = 45 – x – 5

= 40 – x

72 MATHEMATICS

Therefore, their product = (x – 5) (40 – x)

= 40x – x2 – 200 + 5x

= – x2 + 45x – 200

So, – x2 + 45x – 200 = 124 (Given that product = 124)

i.e., – x2 + 45x – 324 = 0

i.e., x2 – 45x + 324 = 0

Therefore, the number of marbles John had, satisfies the quadratic equation

x2 – 45x + 324 = 0

which is the required representation of the problem mathematically.

(ii) Let the number of toys produced on that day be x.

Therefore, the cost of production (in rupees) of each toy that day = 55 – x

So, the total cost of production (in rupees) that day = x (55 – x)

Therefore, x (55 – x) = 750

i.e., 55x – x2 = 750

i.e., – x2 + 55x – 750 = 0

i.e., x2 – 55x + 750 = 0

Therefore, the number of toys produced that day satisfies the quadratic equation

x2 – 55x + 750 = 0

which is the required representation of the problem mathematically.

Example 2 : Check whether the following are quadratic equations:

(i) (x – 2)2 + 1 = 2x – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2)

(iii) x (2x + 3) = x2 + 1 (iv) (x + 2)3 = x3 – 4

Solution :

(i) LHS = (x – 2)2 + 1 = x2 – 4x + 4 + 1 = x2 – 4x + 5

Therefore, (x – 2)2 + 1 = 2x – 3 can be rewritten as

x2 – 4x + 5 = 2x – 3

i.e., x2 – 6x + 8 = 0

It is of the form ax2 + bx + c = 0.

Therefore, the given equation is a quadratic equation.


(ii) Since x(x + 1) + 8 = x2 + x + 8 and (x + 2)(x – 2) = x2 – 4

Therefore, x2 + x + 8 = x2 – 4

i.e., x + 12 = 0

It is not of the form ax2 + bx + c = 0.

Therefore, the given equation is not a quadratic equation.

(iii) Here, LHS = x (2x + 3) = 2x2 + 3x

So, x (2x + 3) = x2 + 1 can be rewritten as

2x2 + 3x = x2 + 1

Therefore, we get x2 + 3x – 1 = 0


So, the given equation is a quadratic equation.

(iv) Here, LHS = (x + 2)3 = x3 + 6x2 + 12x + 8

Therefore, (x + 2)3 = x3 – 4 can be rewritten as

x3 + 6x2 + 12x + 8 = x3 – 4

i.e., 6x2 + 12x + 12 = 0 or, x2 + 2x + 2 = 0


So, the given equation is a quadratic equation.

Remark : Be careful! In (ii) above, the given equation appears to be a quadratic

equation, but it is not a quadratic equation.

In (iv) above, the given equation appears to be a cubic equation (an equation of

degree 3) and not a quadratic equation. But it turns out to be a quadratic equation. As

you can see, often we need to simplify the given equation before deciding whether it

is quadratic or not.

EXERCISE 4.1

1. Check whether the following are quadratic equations :

(i) (x + 1)2 = 2(x – 3) (ii) x2 – 2x = (–2) (3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3) (iv) (x – 3)(2x +1) = x(x + 5)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1) (vi) x2 + 3x + 1 = (x – 2)2

(vii) (x + 2)3 = 2x (x2 – 1) (viii) x3 – 4x2 – x + 1 = (x – 2)3

2. Represent the following situations in the form of quadratic equations :

(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one

more than twice its breadth. We need to find the length and breadth of the plot.

74 MATHEMATICS

(ii) The product of two consecutive positive integers is 306. We need to find the

integers.

(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years)

3 years from now will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been

8 km/h less, then it would have taken 3 hours more to cover the same distance. We

need to find the speed of the train.

4.3 Solution of a Quadratic Equation by Factorisation

Consider the quadratic equation 2x2 – 3x + 1 = 0. If we replace x by 1 on the

LHS of this equation, we get (2 × 12) – (3 × 1) + 1 = 0 = RHS of the equation.

We say that 1 is a root of the quadratic equation 2x2 – 3x + 1 = 0. This also means that

1 is a zero of the quadratic polynomial 2x2 – 3x + 1.

In general, a real number " is called a root of the quadratic equation

ax2 + bx + c = 0, a ! 0 if a "2 + b" + c = 0. We also say that x = """"" is a solution of

the quadratic equation, or that """"" satisfies the quadratic equation. Note that the

zeroes of the quadratic polynomial ax2 + bx + c and the roots of the quadratic

equation ax2 + bx + c = 0 are the same.

You have observed, in Chapter 2, that a quadratic polynomial can have at most

two zeroes. So, any quadratic equation can have atmost two roots.

You have learnt in Class IX, how to factorise quadratic polynomials by splitting

their middle terms. We shall use this knowledge for finding the roots of a quadratic

equation. Let us see how.

Example 3 : Find the roots of the equation 2x2 – 5x + 3 = 0, by factorisation.

Solution : Let us first split the middle term – 5x as –2x –3x [because (–2x) × (–3x) =

6x2 = (2x2) × 3].

So, 2x2 – 5x + 3 = 2x2 – 2x – 3x + 3 = 2x (x – 1) –3(x – 1) = (2x – 3)(x – 1)

Now, 2x2 – 5x + 3 = 0 can be rewritten as (2x – 3)(x – 1) = 0.

So, the values of x for which 2x2 – 5x + 3 = 0 are the same for which (2x – 3)(x – 1) = 0,

i.e., either 2x – 3 = 0 or x – 1 = 0.

Now, 2x – 3 = 0 gives 3

2x ! and x – 1 = 0 gives x = 1.

So,3

2x ! and x = 1 are the solutions of the equation.

In other words, 1 and 3

2 are the roots of the equation 2x2 – 5x + 3 = 0.

Verify that these are the roots of the given equation.


Note that we have found the roots of 2x2 – 5x + 3 = 0 by factorising

2x2 – 5x + 3 into two linear factors and equating each factor to zero.

Example 4 : Find the roots of the quadratic equation 6x2 – x – 2 = 0.

Solution : We have

6x2 – x – 2 = 6x2 + 3x – 4x – 2

= 3x (2x + 1) – 2 (2x + 1)

= (3x – 2)(2x + 1)

The roots of 6x2 – x – 2 = 0 are the values of x for which (3x – 2)(2x + 1) = 0

Therefore, 3x – 2 = 0 or 2x + 1 = 0,

i.e., x =2

3or x =

1

2"

Therefore, the roots of 6x2 – x – 2 = 0 are 2 1

.and –3 2

We verify the roots, by checking that 2 1

and3 2

" satisfy 6x2 – x – 2 = 0.

Example 5 : Find the roots of the quadratic equation 23 2 6 2 0x x" # ! .

Solution : 23 2 6 2x x" # = 23 6 6 2x x x" " #

= $ % $ %3 3 2 2 3 2x x x" " "

= $ %$ %3 2 3 2x x" "

So, the roots of the equation are the values of x for which

$ %$ %3 2 3 2 0x x" " !

Now, 3 2 0x " ! for 2

3x ! .

So, this root is repeated twice, one for each repeated factor 3 2x " .

Therefore, the roots of 23 2 6 2 0x x" # ! are 2

3,

2

3.

76 MATHEMATICS

Example 6 : Find the dimensions of the prayer hall discussed in Section 4.1.

Solution : In Section 4.1, we found that if the breadth of the hall is x m, then x

satisfies the equation 2x2 + x – 300 = 0. Applying the factorisation method, we write

this equation as

2x2 – 24x + 25x – 300 = 0

2x (x – 12) + 25 (x – 12) = 0

i.e., (x – 12)(2x + 25) = 0

So, the roots of the given equation are x = 12 or x = – 12.5. Since x is the breadth

of the hall, it cannot be negative.

Thus, the breadth of the hall is 12 m. Its length = 2x + 1 = 25 m.

EXERCISE 4.2

1. Find the roots of the following quadratic equations by factorisation:

(i) x2 – 3x – 10 = 0 (ii) 2x2 + x – 6 = 0

(iii) 22 7 5 2 0x x# # ! (iv) 2x2 – x +

1

8 = 0

(v) 100 x2 – 20x + 1 = 0

2. Solve the problems given in Example 1.

3. Find two numbers whose sum is 27 and product is 182.

4. Find two consecutive positive integers, sum of whose squares is 365.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find

the other two sides.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed

on a particular day that the cost of production of each article (in rupees) was 3 more than

twice the number of articles produced on that day. If the total cost of production on that

day was Rs 90, find the number of articles produced and the cost of each article.

4.4 Solution of a Quadratic Equation by Completing the Square

In the previous section, you have learnt one method of obtaining the roots of a quadratic

equation. In this section, we shall study another method.

Consider the following situation:

The product of Sunita’s age (in years) two years ago and her age four years

from now is one more than twice her present age. What is her present age?

To answer this, let her present age (in years) be x. Then the product of her ages

two years ago and four years from now is (x – 2)(x + 4).


Therefore, (x – 2)(x + 4) = 2x + 1

i.e., x2 + 2x – 8 = 2x + 1

i.e., x2 – 9 = 0

So, Sunita’s present age satisfies the quadratic equation x2 – 9 = 0.

We can write this as x2 = 9. Taking square roots, we get x = 3 or x = – 3. Since

the age is a positive number, x = 3.

So, Sunita’s present age is 3 years.

Now consider the quadratic equation (x + 2)2 – 9 = 0. To solve it, we can write

it as (x + 2)2 = 9. Taking square roots, we get x + 2 = 3 or x + 2 = – 3.

Therefore, x = 1 or x = –5

So, the roots of the equation (x + 2)2 – 9 = 0 are 1 and – 5.

In both the examples above, the term containing x is completely inside a square,

and we found the roots easily by taking the square roots. But, what happens if we are

asked to solve the equation x2 + 4x – 5 = 0? We would probably apply factorisation to

do so, unless we realise (somehow!) that x2 + 4x – 5 = (x + 2)2 – 9.

So, solving x2 + 4x – 5 = 0 is equivalent to solving (x + 2)2 – 9 = 0, which we have

seen is very quick to do. In fact, we can convert any quadratic equation to the form

(x + a)2 – b2 = 0 and then we can easily find its roots. Let us see if this is possible.

Look at Fig. 4.2.

In this figure, we can see how x2 + 4x is being converted to (x + 2)2 – 4.

Fig. 4.2

78 MATHEMATICS

The process is as follows:

x2 + 4x = (x2 + 4

2x ) +

4

2x

= x2 + 2x + 2x

= (x + 2) x + 2 × x

= (x + 2) x + 2 × x + 2 × 2 – 2 × 2

= (x + 2) x + (x + 2) × 2 – 2 × 2

= (x + 2) (x + 2) – 22

= (x + 2)2 – 4

So, x2 + 4x – 5 = (x + 2)2 – 4 – 5 = (x + 2)2 – 9

So, x2 + 4x – 5 = 0 can be written as (x + 2)2 – 9 = 0 by this process of completing

the square. This is known as the method of completing the square.

In brief, this can be shown as follows:

x2 + 4x =

2 2 24 4 4

42 2 2

x x& ' & ' & '

# " ! # "( ) ( ) ( )* + * + * +

So, x2 + 4x – 5 = 0 can be rewritten as

24

4 52

x& '# " "( )* +

= 0

i.e., (x + 2)2 – 9 = 0

Consider now the equation 3x2 – 5x + 2 = 0. Note that the coefficient of x2 is not

a perfect square. So, we multiply the equation throughout by 3 to get

9x2 – 15x + 6 = 0

Now, 9x2 – 15x + 6 = 2 5(3 ) 2 3 6

2x x" , , #

=

2 2

2 5 5 5(3 ) 2 3 6

2 2 2x x

& ' & '" , , # " #( ) ( )* + * +

=

25 25

3 62 4

x& '" " #( )* +

=

25 1

32 4

x& '" "( )* +


So, 9x2 – 15x + 6 = 0 can be written as

25 1

32 4

x& '" "( )* +

= 0

i.e.,

25

32

x& '"( )* +

=1

4

So, the solutions of 9x2 – 15x + 6 = 0 are the same as those of

25 1

32 4

x& '" !( )* +

.

i.e., 3x – 5

2 =

1

2or

53

2x" =

1

2"

(We can also write this as 5 1

32 2

x " ! - , where ‘#’ denotes ‘plus minus’.)

Thus, 3x =5 1

2 2# or

5 13

2 2x ! "

So, x =5 1

6 6# or

5 1

6 6x ! "

Therefore, x = 1 or x = 4

6

i.e., x = 1 or x = 2

3

Therefore, the roots of the given equation are 1 and 2 .3

Remark : Another way of showing this process is as follows :

The equation 3x2 – 5x + 2 = 0

is the same as

2 5 2

3 3x x" # = 0

Now, x2 – 5 2

3 3x # =

2 21 5 1 5 2

2 3 2 3 3x. / . /& ' & '" " #0 1 0 1( ) ( )

* + * +2 3 2 3

80 MATHEMATICS

=

25 2 25

6 3 36x& '" # "( )* +

=

2 2 25 1 5 1

6 36 6 6x x& ' & ' & '

" " ! " "( ) ( ) ( )* + * + * +

So, the solutions of 3x2 – 5x + 2 = 0 are the same as those of 2 2

5 10

6 6x& ' & '" " !( ) ( )* + * +

,

which are x – 5

6 = ±

1

6, i.e., x =

5 1

6 6# = 1 and x =

5 1

6 6" =

2

3.

Let us consider some examples to illustrate the above process.

Example 7 : Solve the equation given in Example 3 by the method of completing the

square.

Solution : The equation 2x2 – 5x + 3 = 0 is the same as 2 5 30.

2 2x x" # !

Now,2 5 3

2 2x x" # =

2 25 5 3

4 4 2x& ' & '" " #( ) ( )* + * +

=

25 1

4 16x& '" "( )* +

Therefore, 2x2 – 5x + 3 = 0 can be written as

25 1

04 16

x& '" " !( )* +

.

So, the roots of the equation 2x2 – 5x + 3 = 0 are exactly the same as those of

25 1

04 16

x& '" " !( )* +

. Now,

25 1

4 16x& '" "( )* +

=0 is the same as

25 1

4 16x& '" !( )* +

Therefore,5

4x " =

1

4-

i.e., x =5 1

4 4-

i.e., x =5 1 5 1

or4 4 4 4

x# ! "

i.e., x =3

2 or x = 1


Therefore, the solutions of the equations are 3

2x ! and 1.

Let us verify our solutions.

Putting3

2x ! in 2x2 – 5x + 3 = 0, we get

23 3

2 – 5 3 02 2

& ' & ' # !( ) ( )* + * +

, which is

correct. Similarly, you can verify that x = 1 also satisfies the given equation.

In Example 7, we divided the equation 2x2 – 5x + 3 = 0 throughout by 2 to get

x2 – 5 3

2 2x # = 0 to make the first term a perfect square and then completed the

square. Instead, we can multiply throughout by 2 to make the first term as 4x2 = (2x)2

and then complete the square.

This method is illustrated in the next example.

Example 8 : Find the roots of the equation 5x2 – 6x – 2 = 0 by the method of completing

the square.

Solution : Multiplying the equation throughout by 5, we get

25x2 – 30x – 10 = 0

This is the same as

(5x)2 – 2 × (5x) × 3 + 32 – 32 – 10 = 0

i.e., (5x – 3)2 – 9 – 10 = 0

i.e., (5x – 3)2 – 19 = 0

i.e., (5x – 3)2 = 19

i.e., 5x – 3 = 19-

i.e., 5x = 3 19-

So, x =3 19

5

-

Therefore, the roots are 3 19

5

# and

3 19

5

".

Verify that the roots are 3 19

5

# and

3 19

5

".

82 MATHEMATICS

Example 9 : Find the roots of 4x2 + 3x + 5 = 0 by the method of completing the

square.

Solution : Note that 4x2 + 3x + 5 = 0 is the same as

(2x)2 + 2 × (2x) ×

2 23 3 3

54 4 4

& ' & '# " #( ) ( )* + * +

= 0

i.e.,

23 9

2 54 16

x& '# " #( )* +

= 0

i.e.,

23 71

24 16

x& '

# #( )* +

= 0

i.e.,

23

24

x& '

#( )* +

=71

06

"4

But

23

24

x& '#( )* +

cannot be negative for any real value of x (Why?). So, there is

no real value of x satisfying the given equation. Therefore, the given equation has no

real roots.

Now, you have seen several examples of the use of the method of completing

the square. So, let us give this method in general.

Consider the quadratic equation ax2 + bx + c = 0 (a ! 0). Dividing throughout by

a, we get 2 0

b cx x

a a# # !

This is the same as

2 2

02 2

b b cx

a a a

& ' & '# " # !( ) ( )* + * +

i.e.,

2 2

2

4

2 4

b b acx

a a

"& '# "( )* +

= 0

So, the roots of the given equation are the same as those of

2 2

2

40,

2 4

b b acx

a a

"& '# " !( )* +

i.e., those of

2 2

2

4

2 4

b b acx

a a

"& '# !( )* +

(1)


If b2 – 4ac $ 0, then by taking the square roots in (1), we get

2

bx

a# =

2 4

2

b ac

a

- "

Therefore, x =2 4

2

b b ac

a

" - "

So, the roots of ax2 + bx + c = 0 are 2 2

4 4and

2 2

b b ac b b ac

a a

" # " " " ", if

b2 – 4ac $ 0. If b2 – 4ac < 0, the equation will have no real roots. (Why?)

Thus, if b2 – 4ac $%$%$%$%$% 0, then the roots of the quadratic equation

ax2 + bx + c = 0 are given by

2– ± – 4

2

b b ac

a

This formula for finding the roots of a quadratic equation is known as the

quadratic formula.

Let us consider some examples for illustrating the use of the quadratic formula.

Example 10 : Solve Q. 2(i) of Exercise 4.1 by using the quadratic formula.

Solution : Let the breadth of the plot be x metres. Then the length is (2x + 1) metres.

Then we are given that x(2x + 1) = 528, i.e., 2x2 + x – 528 = 0.

This is of the form ax2 + bx + c = 0, where a = 2, b = 1, c = – 528.

So, the quadratic formula gives us the solution as

x =1 1 4(2)(528) 1 4225 1 65

4 4 4

" - # " - " -! !

i.e., x =64 – 66

or4 4

x !

i.e., x = 16 or x = 33

2"

Since x cannot be negative, being a dimension, the breadth of the plot is

16 metres and hence, the length of the plot is 33m.

You should verify that these values satisfy the conditions of the problem.

84 MATHEMATICS

Example 11 : Find two consecutive odd positive integers, sum of whose squares

is 290.

Solution : Let the smaller of the two consecutive odd positive integers be x. Then, the

second integer will be x + 2. According to the question,

x2 + (x + 2)2 = 290

i.e., x2 + x2 + 4x + 4 = 290

i.e., 2x2 + 4x – 286 = 0

i.e., x2 + 2x – 143 = 0

which is a quadratic equation in x.

Using the quadratic formula, we get

x =2 4 572 2 576 2 24

2 2 2

" - # " - " -! !

i.e., x = 11 or x = – 13

But x is given to be an odd positive integer. Therefore, x ! – 13, x = 11.

Thus, the two consecutive odd integers are 11 and 13.

Check : 112 + 132 = 121 + 169 = 290.

Example 12 : A rectangular park is to be designed whose breadth is 3 m less than its

length. Its area is to be 4 square metres more than the area of a park that has already

been made in the shape of an isosceles triangle with its base as the breadth of the

rectangular park and of altitude 12 m (see Fig. 4.3). Find its length and breadth.

Solution : Let the breadth of the rectangular park be x m.

So, its length = (x + 3) m.

Therefore, the area of the rectangular park = x(x + 3) m2 = (x2 + 3x) m2.

Now, base of the isosceles triangle = x m.

Therefore, its area = 1

2 × x × 12 = 6 x m2.

According to our requirements,

x2 + 3x = 6x + 4

i.e., x2 – 3x – 4 = 0

Using the quadratic formula, we getFig. 4.3


x =3 25

2

- =

3 5

2

- = 4 or – 1

But x ! – 1 (Why?). Therefore, x = 4.

So, the breadth of the park = 4m and its length will be 7m.

Verification : Area of rectangular park = 28 m2,

area of triangular park = 24 m2 = (28 – 4) m2

Example 13 : Find the roots of the following quadratic equations, if they exist, using

the quadratic formula:

(i) 3x2 – 5x + 2 = 0 (ii) x2 + 4x + 5 = 0 (iii) 2x2 – 2 2 x + 1 = 0

Solution :

(i) 3x2 – 5x + 2 = 0. Here, a = 3, b = – 5, c = 2. So, b2 – 4ac = 25 – 24 = 1 & 0.

Therefore, x = 5 1 5 1

6 6

- -! , i.e., x = 1 or x =

2

3

So, the roots are 2

3 and 1.

(ii) x2 + 4x + 5 = 0. Here, a = 1, b = 4, c = 5. So, b2 – 4ac = 16 – 20 = – 4 < 0.

Since the square of a real number cannot be negative, therefore 2 4b ac" will

not have any real value.

So, there are no real roots for the given equation.

(iii) 2x2 – 2 2 x + 1 = 0. Here, a = 2, b = 2 2" , c = 1.

So, b2 – 4ac = 8 – 8 = 0

Therefore, x = 2 2 0 2

04 2

-! - ,

1 .i.e.,2

x !

So, the roots are 1

2,

1

2.

86 MATHEMATICS

Example 14 : Find the roots of the following equations:

(i)1

3, 0x xx

# ! 5 (ii)1 1

3, 0,22

xx x" ! 5

"

Solution :

(i)1

3xx

# ! . Multiplying throughout by x, we get

x2 + 1 = 3x

i.e., x2 – 3x + 1 = 0, which is a quadratic equation.

Here, a = 1, b = – 3, c = 1

So, b2 – 4ac = 9 – 4 = 5 > 0

Therefore, x =3 5

2

-(Why?)

So, the roots are 3 5 3 5

and2 2

# ".

(ii)1 1

3, 0, 22

xx x" ! 5

".

As x ! 0, 2, multiplying the equation by x (x – 2), we get

(x – 2) – x = 3x (x – 2)

= 3x2 – 6x

So, the given equation reduces to 3x2 – 6x + 2 = 0, which is a quadratic equation.

Here, a = 3, b = – 6, c = 2. So, b2 – 4ac = 36 – 24 = 12 > 0

Therefore, x =6 12 6 2 3 3 3

.6 6 3

- - -! !

So, the roots are 3 3 3 3

and3 3

# ".


Example 15 : A motor boat whose speed is 18 km/h in still water takes 1 hour more

to go 24 km upstream than to return downstream to the same spot. Find the speed of

the stream.

Solution : Let the speed of the stream be x km/h.

Therefore, the speed of the boat upstream = (18 – x) km/h and the speed of the boat

downstream = (18 + x) km/h.

The time taken to go upstream = distance 24

speed 18 x!

" hours.

Similarly, the time taken to go downstream = 24

18 x# hours.

According to the question,

24 24

18 18x x"

" # = 1

i.e., 24(18 + x) – 24(18 – x) = (18 – x) (18 + x)

i.e., x2 + 48x – 324 = 0

Using the quadratic formula, we get

x =

248 48 1296

2

" - # =

48 3600

2

" -

=48 60

2

" - = 6 or – 54

Since x is the speed of the stream, it cannot be negative. So, we ignore the root

x = – 54. Therefore, x = 6 gives the speed of the stream as 6 km/h.

EXERCISE 4.3

1. Find the roots of the following quadratic equations, if they exist, by the method of

completing the square:

(i) 2x2 – 7x + 3 = 0 (ii) 2x2 + x – 4 = 0

(iii) 24 4 3 3 0x x# # ! (iv) 2x2 + x + 4 = 0

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic

formula.

88 MATHEMATICS

3. Find the roots of the following equations:

(i)1

3, 0x xx

" ! 5 (ii)1 1 11

4 7 30x x" !

# ", x ! – 4, 7

4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from

now is 1

.3

Find his present age.

5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got

2 marks more in Mathematics and 3 marks less in English, the product of their marks

would have been 210. Find her marks in the two subjects.

6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer

side is 30 metres more than the shorter side, find the sides of the field.

7. The difference of squares of two numbers is 180. The square of the smaller number is 8

times the larger number. Find the two numbers.

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would

have taken 1 hour less for the same journey. Find the speed of the train.

9. Two water taps together can fill a tank in 93

8 hours. The tap of larger diameter takes 10

hours less than the smaller one to fill the tank separately. Find the time in which each tap

can separately fill the tank.

10. An express train takes 1 hour less than a passenger train to travel 132 km between

Mysore and Bangalore (without taking into consideration the time they stop at

intermediate stations). If the average speed of the express train is 11km/h more than that

of the passenger train, find the average speed of the two trains.

11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m,

find the sides of the two squares.

4.5 Nature of Roots

In the previous section, you have seen that the roots of the equation ax2 + bx + c = 0

are given by

x =

2– 4

2

b b ac

a

- "

If b2 – 4ac > 0, we get two distinct real roots 2 4

2 2

b acb

a a

"" # and

24

–2 2

b acb

a a

"" .


If b2 – 4ac = 0, then x = 02

b

a" - , i.e., or –

2 2

b bx

a a! " 6

So, the roots of the equation ax2 + bx + c = 0 are both 2

b

a

"6

Therefore, we say that the quadratic equation ax2 + bx + c = 0 has two equal

real roots in this case.

If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac. Therefore,

there are no real roots for the given quadratic equation in this case.

Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has

real roots or not, b2 – 4ac is called the discriminant of this quadratic equation.

So, a quadratic equation ax2 + bx + c = 0 has

(i) two distinct real roots, if b2 – 4ac > 0,

(ii) two equal real roots, if b2 – 4ac = 0,

(iii) no real roots, if b2 – 4ac < 0.

Let us consider some examples.

Example 16 : Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, and

hence find the nature of its roots.

Solution : The given equation is of the form ax2 + bx + c = 0, where a = 2, b = – 4 and

c = 3. Therefore, the discriminant

b2 – 4ac = (– 4)2 – (4 × 2 × 3) = 16 – 24 = – 8 < 0

So, the given equation has no real roots.

Example 17 : A pole has to be erected at a point on the boundary of a circular park

of diameter 13 metres in such a way that the differences of its distances from two

diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to

do so? If yes, at what distances from the two gates should the pole be erected?

Solution : Let us first draw the diagram

(see Fig. 4.4).

Let P be the required location of the

pole. Let the distance of the pole from the

gate B be x m, i.e., BP = x m. Now the

difference of the distances of the pole from

the two gates = AP – BP (or, BP – AP) =

7 m. Therefore, AP = (x + 7) m.

Fig. 4.4

90 MATHEMATICS

Now, AB = 13m, and since AB is a diameter,

'APB = 90° (Why?)

Therefore, AP2 + PB2 = AB2 (By Pythagoras theorem)

i.e., (x + 7)2 + x2 = 132

i.e., x2 + 14x + 49 + x2 = 169

i.e., 2x2 + 14x – 120 = 0

So, the distance ‘x’ of the pole from gate B satisfies the equation

x2 + 7x – 60 = 0

So, it would be possible to place the pole if this equation has real roots. To see if this

is so or not, let us consider its discriminant. The discriminant is

b2 – 4ac = 72 – 4 × 1 × (– 60) = 289 > 0.

So, the given quadratic equation has two real roots, and it is possible to erect the

pole on the boundary of the park.

Solving the quadratic equation x2 + 7x – 60 = 0, by the quadratic formula, we get

x =7 289

2

" - =

7 17

2

" -

Therefore, x = 5 or – 12.

Since x is the distance between the pole and the gate B, it must be positive.

Therefore, x = – 12 will have to be ignored. So, x = 5.

Thus, the pole has to be erected on the boundary of the park at a distance of 5m

from the gate B and 12m from the gate A.

Example 18 : Find the discriminant of the equation 3x2 – 2x +1

3 = 0 and hence find

the nature of its roots. Find them, if they are real.

Solution : Here a = 3, b = – 2 and 1

3c ! .

Therefore, discriminant b2 – 4ac = (– 2)2 – 4 × 3 × 1

3 = 4 – 4 = 0.

Hence, the given quadratic equation has two equal real roots.

The roots are 2 2 1 1, ,, , .i.e., , i.e.,

2 2 6 6 3 3

b b

a a

" "


EXERCISE 4.4

1. Find the nature of the roots of the following quadratic equations. If the real roots exist,

find them:

(i) 2x2 – 3x + 5 = 0 (ii) 3x2 – 4 3 x + 4 = 0

(iii) 2x2– 6x + 3 = 0

2. Find the values of k for each of the following quadratic equations, so that they have two

equal roots.

(i) 2x2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0

3. Is it possible to design a rectangular mango grove whose length is twice its breadth,

and the area is 800 m2? If so, find its length and breadth.

4. Is the following situation possible? If so, determine their present ages.

The sum of the ages of two friends is 20 years. Four years ago, the product of their ages

in years was 48.

5. Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find

its length and breadth.

4.6 Summary

In this chapter, you have studied the following points:

1. A quadratic equation in the variable x is of the form ax2+ bx + c = 0, where a, b, c are real

numbers and a ! 0.

2. A real number " is said to be a root of the quadratic equation ax2 + bx + c = 0, if

a"2 + b" + c = 0. The zeroes of the quadratic polynomial ax2 + bx + c and the roots of the

quadratic equation ax2 + bx + c = 0 are the same.

3. If we can factorise ax2 + bx + c, a ! 0, into a product of two linear factors, then the roots

of the quadratic equation ax2 + bx + c = 0 can be found by equating each factor to zero.

4. A quadratic equation can also be solved by the method of completing the square.

5. Quadratic formula: The roots of a quadratic equation ax2 + bx + c = 0 are given by

24

,2

b b ac

a

" - " provided b2 – 4ac $ 0.

6. A quadratic equation ax2 + bx + c = 0 has

(i) two distinct real roots, if b2 – 4ac > 0,

(ii) two equal roots (i.e., coincident roots), if b2 – 4ac = 0, and

(iii) no real roots, if b2 – 4ac < 0.

92 MATHEMATICS

A NOTE TO THE READER

In case of word problems, the obtained solutions should always be

verified with the conditions of the original problem and not in the

equations formed (see Examples 11, 13, 19 of Chapter 3 and

Examples 10, 11, 12 of Chapter 4).

_________________________________________________________________________________________________

_________________________________________________________________________________________________

www.TCYonline.com Page : 1

®Top Careers & You

Quadratic Equations

(Polynomials, Equations, Remainder theorem)

Polynomials

Definition: Let a0, a1, a2, ….. an be real numbers and x is a real variable. Then f(x) = a0 + a1 x + a2 x2 + ……..

+ anxn is called a real polynomial of real variable x with coefficients a0, a1, a2, ….. an.

Examples: x3 + 4x

2 – 3 is a polynomial

Degree of a Polynomial:

Degree of a polynomial is the highest power of the variable in the polynomial.

Example:

Degree of 3x3 + 4x

2 – 3 is 3 as the maximum power of the variable x is 3.

On the basis of degree, the polynomials are classified as Linear (Degree 1), Quadratic (degree 2), Cubical

(Degree 3), bi-Quadratic (degree 4), and so on.

Remainder Theorem

If any polynomial f(x) is divided by (x – a) then f(a) is the remainder.

For example,

f(x) = x2 – 5x + 7 = 0 is divided by x – 2. What is the remainder?

R = f(2) = 22 – 5 ! 2 + 7 = 1.

Factor Theorem

If (x – a) is a factor of f(x), then remainder f(a) = 0. (Or) if f(a) = 0, then (x – a) is a factor of f(x),

For example,

When f(x) = x2 – 5x + 6 = 0 is divided by x – 2, the remainder f(2) is zero which shows that x – 2 is the factor of

f(x)

General Theory of Equations

An equation is the form of a polynomial which has been equated to some real value.

For example:

2x + 5 = 0, x2 – 2x + 5 = 7, 2x

2 – 5x

2 + 1 = 2x + 5 etc. are polynomial equations.

Root or Zero of a polynomial equation:

If f (x) = 0 is a polynomial equation and f (" ) = 0, then " is called a root or zero of the polynomial equation

f(x) = 0.

Linear Equation

Linear Equation with one variable:

A linear equation is 1st

degree equation. It has only one root. Its general form is a

x + b = 0 and root is –a

b.

TIP

If we plot the graph

of y = f(x) A linear

equation is 1st degree

equation. It has only

one root. Its general

form is a x + b = 0 and

root is – a

b.

_________________________________________________________________________________________________

_________________________________________________________________________________________________


®Top Careers & You

Linear equation with two variables:

It is a first degree equation with two variables.

Ex: 2x + 3y = 0.

We need two equations to find the values of x and y.

If there are n variables in an equation, we need n equations to find the values of the variables uniquely.

Some times, even the number of equations are equal to the number of variables, we cannot find the values of

x and y uniquely.

For example, 3x + 5y = 6

6x +10y = 12

These are two equations, but both are one and the same. So different values of x and y satisfy the equation

and there is no unique solution. It will has infinite number of solutions.

The number of solutions is clearly described below for the set of equations with 2 variables.

a1x + b1y + c1 = 0

a2x + b2y + c2 = 0

These equations can be

1. Inconsistent means have no solution if 2

1

2

1

2

1

c

c

b

b

a

a#$

2. Consistent and has infinitely many solutions if2

1

2

1

2

1

c

c

b

b

a

a$$

3. Consistent and have unique solution if 2

1

2

1

b

b

a

a#

Quadratic Equation

Quadratic Equation in “x” is one in which the highest power of “x” is 2. The equation is generally satisfied by

two values of “x”.

The quadratic form is generally represented by ax2 + bx + c = 0 where a # 0, and a, b, c are constants.

For Example:

x2 – 6x + 4 = 0

3x2 + 7x – 2 = 0

A quadratic equation in one variable has two and only two roots, which are

a2

ac4bbx

2

1

%&%$ &

a2

ac4bbx

2

2

%%%$

Nature of Roots

The two roots of any quadratic equation always depend on the value of

b2 – 4ac called discriminant (D).

D > 0 Real and unequal roots

D = 0 Real and equal

D < 0 Imaginary and unequal

F U N D A

Imaginary roots are

always unequal and

conjugate of each other.

i.e. If one root is 3 + 5i

the other will be 3 – 5i.

T I P S

1. If roots of given

equation are equal in

magnitude but opposite

in sign, then b = 0 & vice

versa.

2. If roots of given

equation are reciprocal

of each other, then c =

a.

_________________________________________________________________________________________________

_________________________________________________________________________________________________


®Top Careers & You

Sum and Product of roots:

If '" and are the two roots of ax2 + bx + c = 0,

Then sum of roots = '" & = – a

b = –

)tcoefficienx(

)tcoefficienx(2

And product of roots = "' = a

c=

tcoefficienx

ttancons2

Formation of equation from roots:

If '" and are the roots of any quadratic equation then that equation can be written in the form

( ) 0XX 2 $"'&'&"%

i.e. X2

– (sum of the roots) X + Product of the roots = 0

Some Important results

If x1, x2 are the roots of the equation f(x) = ax2 + bx + c = 0, then,

1. The equation whose roots are equal in magnitude and opposite in

sign to that of f(x) i.e., – x1, – x2 is f(– x) = 0.i.e., ax2 – bx + c = 0.

2. The equation whose roots are reciprocals to that of f(x) i.e.,

21 x

1and

x

1 is f(1/x) = 0. i.e. cx

2 + bx + a = 0.

3. The equation whose roots are k more/less than that of f(x) i.e., x1 * k,

x2 * k is f(x ! k) = 0. i.e. a (x ! k). 2 + b(x ! k) + c = 0.

4. The equation whose roots are k times to that of f(x) i.e., k x1, kx2, is

f(x/k) = 0. i.e. ax2 + kbx + k 2 c = 0

5. The equation whose roots are x1k , x2

k is f( k x ) = 0. i.e.

a ( k x ) 2 + b( k x ) + c = 0.

Relation between roots and coefficient of an equation

Let " 1' " 2' " 3' ……., " n be the n roots of the equation:

a0 xn + a1 x

n–1 + a2 x

n–1 …. + an–1 x + an = 0

Then we have the following relations:

Sum of the roots taken one at a time (" 1 + " 2 +…….+" n) = – (a1/a0).

Sum of the roots taken two at a time (" 1" 2+" 2 " 3 +……." n" 1 ) = (a2/a0)

Sum of the roots taken three at a time (" 1" 2" 3 +" 2" 3" 4 +….….+ " n" 1 " 2) = – (a3/a0)

-----------------------------------------------

Product of the roots = (" 1 " 2 " 3. ………" n) = {(– 1)n an / a0}.

E.g. Polynomial equation ax4 + bx

3 + cx

2 + dx + e = 0

Sum of the roots = – a

b.

Sum of the roots taking two at a time = a

c

Sum of the roots taking three at a time = – a

d

and Product of all the roots =a

e.

Do you know?

1. If one of the

roots of a quadratic

equation is k1 + +k2,

then the other root

will be k1 – +k2 & vice

versa.

2. If one of the

roots be (m + in),

then the other root

will be (m – in) & vice

versa.

_________________________________________________________________________________________________

_________________________________________________________________________________________________


®Top Careers & You

Maximum and Minimum value of a Quadratic equation

The quadratic equation ax2 + bx + c = 0 will have maximum or minimum value at x = – b/2a. If a < 0, it has

maximum value and if a > 0, it has minimum value.

The maximum or minimum value is given by a4

bac4 2%.

Ex.1 Solve for x: 3x

4x3

%&

= 3

2

Sol. 3(3x + 4) = 2(x – 3)

, 7x = – 18 , x = – 7

18

Ex.2 A and B went to a hotel paid Rs. 84 for 3 plates of Idli and 5 plates of Dosa. Where as B took 5

plates of Idli and 3 plates of Dosa and paid Rs. 76. What is the cost of one plate of Idli.

Sol. 3I + 5D = 84 ……….(1)

5I + 3D = 76 ………(2)

Equation (1) ! 3 - equation (2) ! 5, we get

16I = 128

, I = 8

Each plate of Idli cost Rs. 8.

Ex.3 Find the values of x and y from the equations x

4 –

y

3 = 1 and

x

1 +

y

9 = 3.5.

Sol. Take x

1 = a,

y

1 = b

Then, the equations will become

4a – 3b = 1 ……..(1)

And a + 9b = 3.5 ………(2)

(1) ! 3 + (2)

, a = 2

1 and by substituting a in either (1) or (2), we can get b =

3

1.

- a = x

1 =

2

1, x = 4 and b =

y

1 =

3

1, y = 9

Ex.4 Find x and y from yx

1

%+

y2x3

5

% = 2, and

y2x3

15

% –

yx

2

% = 1.

Sol. Take yx

1

% = a and

y2x3

1

% = b

- a + 5b = 2 ……(1)

and 15b – 2a = 1 ……(2)

(1) ! 2 + (2) , b = 5

1

So, a = 1.

_________________________________________________________________________________________________

_________________________________________________________________________________________________


®Top Careers & You

- x – y = a

1 = 1 ……..(3)

and 3x – 2y = b

1 = 5 ……(4)

(3) ! (3) – (4) gives Y = 2 and X = 3.

Ex.5 Aman won a competition and so he got some prize money. He gave Rs. 2000 less than the half

of prize money to his son and Rs. 1000 more than the two third of the remaining to his daughter.

If both they got same amount, what is the prize money Aman got?

Sol. Assume Aman got x rupees.

He gave 2

x – 2000 to his son.

And 3

2./

012

3& 2000

2

x + 1000 to his daughter.

-2

x – 2000 =

3

x +

3

7000

, x = Rs. 26000

Ex.6 How many non negative integer pairs (x, y) satisfy the equation, 3x + 4y = 21?

Sol. Since x and y are non negative integers.

Start from x = 0.

If x = 0 or 2, y cannot be integer.

For x = 3, y = 3.

And for x = 7, y = 0.

These two pairs only satisfy the given equation.

Ex.7 If (x – 2) is a factor of x3 – 3x

2 + px + 4. Find the value of p.

Sol. Since (x – 2) is a factor, f(2) = 0.

- 23 – 3(2

2) + (2)p + 4 = 0

, p = 0

Ex.8 When x3 – 7x

2 + 3x - P is divided by x + 3, the remainder is 4, then what is the value of P ?

Sol. f(– 3) = 4

- (– 3)3 – 7(– 3)

2 + 3(– 3) – P = 4

P = – 103.

Ex.9 If (x – 1) is the HCF of (x3 – px

2 + qx – 3) and (x

3 – 2x

2 + px + 2). What is the value of ‘q’?

Sol. Since (x – 1) is HCF, it is a factor for both the polynomials.

- 13 – p(1)

2 + q(1) – 3 = 0 , – p + q = 2

And 13 – 2(1

2) + p(1) + 2 = 0

p = – 1

- q = 1

_________________________________________________________________________________________________

_________________________________________________________________________________________________


®Top Careers & You

Ex.10 Find the roots of the quadratic equation x2 – x – 12 = 0.

Sol. x2 – x – 12 = 0

, x2 – 4x + 3x – 12 = 0

x(x – 4) – 3(x – 4) = 0

(x – 4) (x – 3) = 0

, x = 4 or 3. The roots are 4 and 3.

Ex.11 Find the roots of the quadratic equation x2 – 8x + 5 = 0.

Sol. x = a2

ac4bb 2 %*%

y = 12

5148)8( 2

!!!%*%%

= 2

448 *= 4 * 11 The roots are 4 + 11 and 4 – 11 .

Ex.12 If 3 + 4i is a root of quadratic equation x2 – px + q = 0. What is the value of pq? (Given i is known

as iota and i2 = –1)

Sol. If 3 + 4i is one root of a quadratic equation, 3 – 4i will be the other root.

(Imaginary roots exist in conjugate pairs)

Sum of roots = p = (3 + 4i) + (3 – 4i) , p = 6

Product of the roots = q = (3 + 4i) (3 – 4i)

, q = 25

- pq = 150.

Ex.13 If one root of a quadratic equation x2 – px + 8 = 0 is square of the other, what is the value of p?

Sol. Let the roots of ", "2.

Sum = " + "2 = p

Product = "("2) = 8 , " = 2

- p = 2 + 22 = 6.

Ex.14 Describe the nature of the roots of the equation x – x

1 = 3.

Sol. Given equation can be written as x2 – 3x – 1 = 0.

Discriminent = (– 3)2 – 4(1) (– 1) = 13 > 0.

So, roots are real and distinct.

Ex.15 If ", ' are the roots of x2 – 7x + P = 0, and " – ' = 3, then what is the value of P?

Sol. Sum = " + ' = 7 ……….(1)

Product = "' = P ……….(2)

Given, " – ' = 3 ………..(3)

(1) and (3) , " = 5, ' = 2

- P = 10 (from (2))

_________________________________________________________________________________________________

_________________________________________________________________________________________________


®Top Careers & You

Ex.16 Form a quadratic equation, whose one of the roots is 2 + 3 .

Sol. If 2 + 3 is one root, other root will be 2 – 3 .

- The equation is x2 – (2 + 3 + 2 – 3 )x + (2 + 3 ) (2 – 3 ) = 0

, x2 – 4x + 1 = 0

Ex.17 Find the values of x, which satisfy the equation 4x – (10) 2

x + 16 = 0.

Sol. Assume 2x = k

- The given equation will become k2 – 10k + 16 = 0.

k2 – 8k – 2k + 16 = 0

(k – 2) (k – 8) = 0

, k = 2 or 8

- 2x = 2 or 2

x = 8

- x = 1 or 3.

Ex.18 What values of x satisfy the given equations 4k.x – x2 = 4k

2 and

(x – k)2 + 3(x – k) – 7 = 0?

Sol. The first equation can be written as (x – 2k)2 = 0.

, x = 2k , k = 2

x

- The second equation will become

2

2

xx .

/

012

3% + 3 .

/

012

3%

2

xx – 7 = 0

x2 + 6x + 28 = 0

- x = 2

112366 %*%

x = – 3 * 37 .

Ex.19 If x1 and x2 are the roots of the equation x2 – 2x + 14 = 0, what is the equation with roots 3x1 – 2

and 3x2 – 2?

Sol. The equation with roots 3x1 and 3x2 is

2

3

x./

012

3 – 2 .

/

012

33

x + 4 = 0.

, x2 – 6x + 36 = 0

The equation with roots 3x1 – 2 and 3x2 – 2 is

(x + 2)2 – 6(x + 2) + 36 = 0

, x2 – 2x + 28 = 0.

Ex.20 If ", ' and 4 are the roots of a cubic equation, x3 – 2x

2 + x – 5 = 0, then what is the value of "2

+ '2

+ 42?

Sol. Sum of roots = " + ' + 4 = 2.

"' + '4 + 4" = 1

"2 + '2

+ 42 = (" + ' + 4)2

– 2("' + '4 + 4")

= 22 – 2(1) = 2.

- "2 + '2

+ 42 = 2.

_________________________________________________________________________________________________

_________________________________________________________________________________________________


®Top Careers & You

Ex.21 Find the value of

....3

12

13

12

13

12

1

&%

&%

&%

.

Sol. Assume the given is x.

So it can be written as x =

x3

12

1

&%

- x = x25

x3

&&

, 5x + 2x2 = 3 + x

2x2 + 4x – 3 = 0

x = 4

24164 &*%

x = – 1 *2

5.

But x cannot be negative.

So x = – 1 + 2

5.

Ex.22 What is the value of x2 – 2x + 3, when the value of 2x

2 – 5x + 6 is minimum?

Sol. The value of 2x2 – 5x + 6 will be minimum at x =

22

)5(

!%%

=4

5./

012

3 %$

a2

bx

- At x = 4

5 x

2 – 2x + 3 =

16

33.

Ex.23 What is the value of k, if the quadratic equation x2 – 6x + k has only one real root?

Sol. If one root is imaginary, the other should be its conjugate, which is also imaginary.

So, the given equation has only one real root and there is no chance for the other root to be imaginary,

so the roots should be equal.

- Discriminent = 0

(– 6)2 – 4 ! 1 ! k = 0

, k = 9.

Ex.24 What is the value of p – q, if the roots of the quadratic equation px2 – (p + q)x + q = 0 are

reciprocals?

Sol. Assume the roots be ","1

.

- Product = p

q = 1 , q = p - p – q = 0

Home Forums CAT 2011 Quant Lessons Theory of Equations..Search forums

< Jump to... >

New Batches at TathaGat Delhi and Pune!--Number System E-Book Updated!Display replies flat, with oldest first

Theory of Equations..by Total Gadha - Saturday, 3 March 2007, 03:06 PM

GENERAL EQUATION OF Nth DEGREE

Let polynomial f(x) = a0xn + a1xn - 1 + a2xn - 2 + ... + an. where a0, a1, a2, ..an are rational numbers and n > 0. Then the values of x for which f(x) reduces

to zero are called root of the equation f(x) = 0. The highest whole number power of x is called the degree of the equation.

For example

x4 - 3x3 + 4x2 + x + 1 = 0 is an equation with degree four.

x5 - 6x4 + 3x2 + 1 = 0 is an equation with degree five.

ax + b = 0 is called the linear equation.

ax2 + bx + c = 0 is called the quadratic equation.

ax3 + bx2 + cx + d = 0 is called the cubic equation.

Properties of equations and their roots

Every equation of the nth degree has exactly n roots.

For example, the equation x3 + 4x2 + 1 = 0 has 3 roots,

The equation x5 - x + 2 = 0 has 5 roots, and so onâ€¦

In an equation with real coefficients imaginary roots occur in pairs i.e. if a + ib is a root of the equation f(x) = 0, then a - ib will also be a root of thesame equation. For example, if 2 + 3i is a root of equation f(x) = 0, 2 - 3i is also a root.

If the coefficients of an equation are all positive then the equation has no positive root. Hence, the equation 2x4 + 3x2 + 5x + 1 = 0 has no positiveroot.If the coefficients of even powers of x are all of one sign, and the coefficients of the odd powers are all of opposite sign, then the equation has no

negative root. Hence, the equation 6x4 - 11x3 + 5x2 - 2x + 1 = 0 has no negative root

If the equation contains only even powers of x and the coefficients are all of the same sign, the equation has no real root. Hence, the equation 4x4 +

5x2 + 2 = 0 has no real root.If the equation contains only odd powers of x, and the coefficients are all of the same sign, the equation has no real root except x = 0. Hence, the

equation 5x5 + 4x3 + x = 0 has only one real root at x = 0.Descartes' Rule of Signs : An equation f(x) = 0 cannot have more positive roots than there are changes of sign in f(x), and cannot have more

negative roots than there changes of sign in f( - x). Thus the equation x4 + 7x3 - 4x2 - x - 7 = 0 has one positive root because there is only change

in sign. f( - x) = x4 - 7x3 - 4x2 + x - 7 = 0 hence the number of negative real roots will be either 1 or 3.

EXAMPLES:

I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some problems based on thisin the CBT Club this week.

If you think this article was useful, help others by sharing it with your friends!

You might also like:Time, Speed and DistanceLinear Equations

Reply

About Contact SMS Updates

Quadratic Equations

Quadratic Equation

An equation of the form ax2 + bx + c = 0, where a 0, is called a quadratic equation. The numbers a, b, c are

called the coefficients of the quadratic equation. A root of the quadratic

equation is a number a (real or complex) such that a 2 + b + c = 0.

The roots of the given quadratic equation are given by x = .

Basic Results:

The quantity D (D = b2 - 4ac) is known as the discriminant of the quadratic equation. For a, b, c real,

The quadratic equation has real and equal roots if and only if D = 0 i.e. b2 - 4ac = 0.

The quadratic equation has real and distinct roots if and only if D > 0 i.e. b2 - 4ac > 0.

The quadratic equation has complex roots with non-zero imaginary parts if and only if D < 0 i.e. b2 - 4ac < 0.

If p + iq (p and q being real) is a root of the quadratic equation where i = , then p - iq is also a root of

the quadratic equation.

If p + is an irrational root of the quadratic equation, then p - is also a root of the quadratic

equation provided that all the coefficients are rational.

The quadratic equation has rational roots if D is a perfect square and a, b, c are rational. If a = 1 and b, c are

integers and the roots of the quadratic equation are rational, then the

roots must be integers.

Factorization Method:

In the factorization method we write the coefficient of x as a sum of two numbers l and m such that lm = ac i.e. we

Dieon Sovban tojo

Zargar Anit Cool

AIEEE on Facebook

61 people like AIEEE.

Like

Facebook social plugin

Online

Editor's Pick

Archives

Tips to Attempt AIEEE

Books For AIEEE

AIEEE Over Last 5 Years

AIEEE Concepts®A Complete Coverage Over AIEEE Exam

Explore Definite Integrals! Click

write

b = l + m with lm = ac

so that the equation ax2 + bx + c = 0 becomes

ax2 + (l + m) x + or a2x2 + a (l + m) x + lm = 0

or (ax + l) (ax + m) = 0 ax + l = 0 or ax + m = 0.

Hence the roots are x = and x = . In order to obtain l and m, we write

l - m = .

With l + m = b, we get

l = and m =

so that the two roots are .

Let and be two roots of the given quadratic equation. Then + = - and = .

A quadratic equation, whose roots are a and b can be written as (x - ) (x - ) = 0

i.e., ax2 + bx + c a(x - ) (x - ).

Condition for two Quadratic Equations to have a Common Root

Let ax2+bx+c = 0 and dx2+ex + f = 0 have a common root (say). Then a 2 + b + c = 0 and d 2 + e + f

= 0.

Solving for 2 and , we get

i.e. 2 = and = (dc - af)2 = (bf - ce) (ae - bd),

which is the required condition for the two equations to have a common root.

Relation between the Roots of a Polynomial Equation of Degree n Consider the equation

anxn + an - 1xn - 1 + an - 2x

n - 2 + .... + a1x + a0 = 0 . . . . (1)

(a0, a1...., an are real coefficients and an 0).

Let 1, 2,...., n be the roots of equation (1). Then

anxn + an - 1xn - 1 + an - 2x

n - 2 + ..... + a1x + a0 an(x - 1) (x - 2) ..... (x - n).

Comparing the coefficients of like powers of x, we get

About AIEEE Exam

AIEEE 2010 Solved Paper







Maths

Sets, Relation & Function

Trigonometry

Mathematical Induction

Complex Numbers

Quadratic Equations

Linear Inequalities

Permutations & Combinations

Binomial Theorem

Sequence & Series

Straight Lines

Conic Section

Limits & Derivatives

Mathematical Reasoning

Statistics

Solutions Of Triangles

Relations & Functions

Inverse Trignometric Functions

Matrices

Determinants

Continuity & Differentiability

Applications Of Derivaties

Indefinite Integeral

Definite Integeral

Differential Equations

Vectors

Probability

3D-Geometry

Physics

Unit & Dimension

Vector Analysis

One Dimensional Motion

Circular Motion

Projectile Motion

Newton's Laws Of Motion

Rotational Motion

Work, Power & Energy

Centre Of Mass

Gravitation

Properties Of Matter

Thermometry

Thermodynamics

Heat Transfer

a1 + 2 + 3 + .... + n = - ,

1a 2 + 1 3 + 1 4 + .... + 2 3 + ... + n - 1 n = ,

1 2 . . . . . r + .... + n-r+1 n-r+2 ... n = ( -1)r ,

1 2 ... n = (-1)n .

Note:

A polynomial equation of degree n has n roots (real or imaginary).

If all the coefficients are real then the imaginary roots occur in pairs i.e. number of complex roots is always

even.

If the degree of a polynomial equation is odd then the number of real roots will also be odd. It follows that at

least one of the roots will be real.

If is a repeated root (repeating r times) of a polynomial equation f(x) = 0 of degree n i.e. f(x) = (x - )r

g(x) , where g(x) is a polynomial of degree n - r and g( ) 0, then f( )

= f'( ) = f''( ) = . . . . = f (r-1)( ) = 0 and f r ( ) 0 .

Remainder Theorem: If we divide a polynomial p(x) by x - , the remainder obtained is p( ). Note that if

p( ) = 0, then x - is a factor of p(x).

If a polynomial equation of degree n has n + 1 roots say x1,...xn + 1, (xi xj if i j), then the polynomial is

identically zero. ie. p(x) = 0, x R.

(In other words, the coefficients a0, .... an are all zero).

If p(a) and p(b) (a < b) are of opposite sign, then p(x) = 0 has odd number of roots in (a, b), i.e. it have at least

one root in (a, b).

If coefficients in p(x) have 'm' changes in signs, then p(x) = 0 have at most 'm' positive real roots and if p(-x)

have 't' changes in sign, then p(x) = 0 have at most 't' negative real roots.

By this we can find maximum number of real roots and minimum number of complex roots of a polynomial equations

p(x) = 0.

The Method of Intervals (Wavy Curve Method)

The Method of intervals (or wavy curve method) is used for solving inequalities of the form

f(x) = > 0 ( < 0, 0, or 0) where n1, n2,..., nk , m1, m2, ...mp

are natural numbers and the numbers a1, a2, ... , ak ; b1, b2,...bp are any real

numbers such that aI bj , where i = 1, 2, 3,..., k and j = 1, 2, 3,..., p .

It consists of the following steps:

All zeros1 of the function f(x) contained on the left hand side of the inequality should be marked on the number

line with inked (black) circles.

All points of discontinuities2 of the function f(x) contained on the left hand side of the inequality should be

Simple Harmonic Motion

Waves

Electrostatics

Current Electricity

Thermal & Chemical Effect Of

Current

Magnetic Effects Of Current &

Magnetism

Magnetism

Magnetic Flux

Alternating Current

Electromagnetic Waves

Ray & Wave Optics

Atomic Physics

Radioactivity

Electronic & Semiconductor

Devices

Communication

Chemistry

Some Basic Concepts Of

Chemistry

Structure Of Atom

Periodic Table & Properties

Chemical Bonding & Molecular

Structure

States Of Matter

Thermodynamics

Chemical Equilibrium

Redox Reactions

Hydrogen

S Block Elements

P Block Elements

Nomenclature & Isomerism Of

Organic Compounds

Hydrocarbons

Environmental Chemistry

Solid State

Solutions

Electrochemistry

Chemical Kinetics

Suface Chemistry & Catalysts

General Principles &

Metallurgy

P Block Elements

D & F Block Elements

Cordination Compounds

Haloalkenes & Haloarenes

Alcohos, Phenols & Ethers

Aldehydes & Ketones

Carboxylic Acids

Amines

Biomolecules & Biological

Processes

Polymers

Chemistry In Everyday Life

marked on the number line with un-inked (white) circles.

Check the value of f(x) for any real number greater than the right most marked number on the number line.

From right to left, beginning above the number line (in case the value of f(x) is positive in step (iii), otherwise,

from below the number line), a wavy curve should be drawn to pass

through all the marked points so that when it passes through a simple point3, the curve intersects the number line,

and, when passing through a double point4, the curve remains located

on one side of the number line.

The appropriate intervals are chosen in accordance with the sign of inequality (the function f(x) is positive

whenever the curve is situated above the number line, it is negative if the

curve is found below the number line). Their union represents the solution of the given inequality.

Remark:

(i) Points of discontinuity will never be included in the answer.

(ii) If you are asked to find the intervals where f(x) is non-negative or non-positive then make the intervals closed

corresponding to the roots of the numerator and let it remain open

corresponding to the roots of the denominator.

1.The point for which f(x) vanishes (becomes zero) are called function zeros e.g. x = ai.

2. The points x = bj are the point of the discontinuity of the function f(x).

3. If the exponents of a factor is odd then the point is called a simple point.

4. If the exponent of a factor is even then the point is called a double point.

Quadratic Expression

The expression ax2 + bx + c is said to be a real quadratic expression in x where a, b, c are real and a 0. Let f(x)

= ax2 + bx + c where a, b, c, R (a 0).

f(x) can be rewritten as f(x) = a =a , where D = b2-4ac is

the discrimnant of the quadratic expression. Then y = f(x) represents a

parabola whose axis is parallel to the y - axis, with vertex at A .

Depending on the sign of a and b2 - 4ac, f(x) may be positive, negative or zero. This gives rise to the following cases:

(i) a > 0 and b2- 4ac < 0

f(x) > 0 x R.

In this case the parabola always remains above the x - axis

.

(ii) a > 0 and b2 - 4ac = 0

f(x) 0 x R.

In this case the parabola touches the

x - axis at one point and remains concave upwards

.

(iii) a > 0 and b2- 4ac > 0.

Let f(x) = 0 have two real roots and ( < ).

Then f(x) > 0 x (- , ) ( , ),

and f(x) < 0 x ( , )

.

(iv) a < 0 and b2 - 4ac < 0

f(x) < 0 x R.

In this case the parabola always remains below the x-axis

.

(v) a < 0 and b2 - 4ac = 0

f(x) 0 x R.

In this case the parabola touches the

x - axis and lies below the x - axis.

(vi) a < 0 and b2- 4ac > 0

Let f(x) = 0 have two real roots a and

( < ). Then f(x) < 0 x (- , ) ( , )

and f(x) > 0 x ( , )

.

Interval in Which the Roots Lie

In some problems we want the roots of the equation ax2 + bx + c = 0 to lie in a given interval. For this we impose

conditions on a, b and c. Let f(x) = ax2 + bx + c.

If both the roots are positive i.e. they lie in (0, ), then the sum of the roots as well as the product of the

roots must be positive

+ = - and = with 2 - 4ac 0.

Similarly, if both the roots are negative i.e. they lie in (- , 0) then the sum of the roots will be negative and the

product of the roots must be positive

i.e. + = - < 0 and = with 2 - 4ac 0.

Both the roots are greater than a given number k if the following three conditions are satisfied: D 0, -

and a.f(k) > 0.

Both the roots will be less than a given number k if the following conditions are satisfied: D 0, - < k

and a.f(k) > 0.

Both the roots will lie in the given interval (k1, k2) if the following conditions are satisfied: D 0 k1 < -

and a. f(k1) > 0, a.f(k2) > 0.

Exactly one of the roots lies in the given interval (k1, k2) if f(k1) . f(k2) < 0.

A given number k will lie between the roots if a.f(k) < 0.

In particular, the roots of the equation will be of opposite signs if 0 lies between the roots a.f(0) < 0. It also

implies that the product of the roots is negative.

Cube Root of Unity:

Consider the equation x3 = 1 x3 - 1 = 0. We note that x = 1 is one of the roots of this equation, so that

0 = x3 - 1 = (x - 1) (x2 + x + 1)

x = 1 or x2 + x + 1 = 0 x = .

Hence the cube roots of unity are: x = 1, x = - + , x = - - .

Newer Post Older Post

Powered By

Alternatively, x = 1, , 2 where = .

Translate:

Read In Hindi

TOP^

75 people onlineThey can help you right now!

Search

powered by FreeFind

Return to the Lessons Index | Do the Lessons in Order | Get "Purplemath on CD" for offl ine use | Print-friendly page

Graphing Quadratic Functions: The Leading Coefficient / The Vertex (page 2 of 4)

Sections: Introduction, The meaning of the leading coefficient / The vertex, Examples

The general form of a quadratic is "y = ax2 + bx + c". For graphing, the leading coefficient "a" indicates

how "fat" or how "skinny" the parabola will be.

For | a | > 1 (such as a = 3 or a = –4), the parabola will be "skinny", because it grows more quickly

(three times as fast or four times as fast, respectively, in the case of our sample values of a). For

| a | < 1 (such as a = 1/3 or a = –1/4 ), the parabola will be "fat", because it grows more slowly (one-

third as fast or one-fourth as fast, respectively, in the examples). Also, if a is negative, then the parabola

is upside-down.

You can see these trends when you look at

how the curve y = ax2 moves as "a"

changes:

As you can see, as the leading coefficient goes from very negative to slightly negative to zero (not reallya quadratic) to slightly positive to very positive, the parabola goes from skinny upside-down to fatupside-down to a straight line (called a "degenerate" parabola) to a fat right-side-up to a skinny right-

side-up. Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

There is a simple, if slightly "dumb", way to remember the difference between right-side-up parabolasand upside-down parabolas:

positive quadratic y = x2 negative quadratic y = –x2

Preparing for IIT?Two Years Course forIIT-JEE 2013Scholarship Test.Enroll Today!www.AakashIITJEE.com

IIT JAM 2012CoachingVPM Classes - JAM2012, 2013, 2014;JAM 2011- 110ranks(30 in top 100)www.vpmclasses.com

CAT 2011 NewFormatiCATs replicate 2 70min sections.Experience the mostrealistic mockswww.TestFunda.com

NATA & AIEEEB.Arch. 2012Buy easily NATA &AIEEE B.Arch 2012Study material + NATAsample paperswww.architectureaptitude.…

IIT JEE ExamPreparationUnlimited Tests.Detailed Analysis ofStrong & Weak Areasonly Rs 550!www.TCYOnline.com/IIT-J…

Purplemath:

Linking to this site

Printing pages

Donating

School l icensing

Reviews of

Internet Sites:

Free Help

Practice

Et Cetera

The "Homework

Guidelines"

Study Skil ls Survey

Tutoring ($$)

This lesson may be printed out for

your personal use.

This can be useful information: If, for instance, you have an equation where a is negative, but you're

somehow coming up with plot points that make it look like the quadratic is right-side-up, then you willknow that you need to go back and check your work, because something is wrong.

Parabolas always have a lowest point (or a highest point, if the parabola is upside-down). This point,where the parabola changes direction, is called the "vertex".

If the quadratic is written in the form y = a(x – h)2 + k, then the vertex is the point (h, k). This makes

sense, if you think about it. The squared part is always positive (for a right-side-up parabola), unless it's

zero. So you'll always have that fixed value k, and then you'll always be adding something to it to make

y bigger, unless of course the squared part is zero. So the smallest y can possibly be is y = k, and this

smallest value will happen when the squared part, x – h, equals zero. And the squared part is zero

when x – h = 0, or when x = h. The same reasoning works, with k being the largest value and the

squared part always subtracting from it, for upside-down parabolas.

(Note: The "a" in the vertex form "y = a(x – h)2 + k" of the quadratic is the same as the "a" in the

common form of the quadratic equation, "y = ax2 + bx + c".)

Since the vertex is a useful point, and since you can "read off" the coordinates for the vertex from thevertex form of the quadratic, you can see where the vertex form of the quadratic can be helpful,especially if the vertex isn't one of your T-chart values. However, quadratics are not usually written in

vertex form. You can complete the square to convert ax2 + bx + c to vertex form, but, for finding the

vertex, it's simpler to just use a formula. (The vertex formula is derived from the completing-the-squareprocess, just as is the Quadratic Formula. In each case, memorization is probably simpler thancompleting the square.)

For a given quadratic y = ax2 + bx + c, the vertex (h, k) is found by computing h = –b/2a, and then

evaluating y at h to find k. If you've already learned the Quadratic Formula, you may find it easy to

memorize the formula for k, since it is related to both the formula for h and the discriminant in the

Quadratic Formula: k = (4ac – b2) / 4a.

Find the vertex of y = 3x2 + x – 2 and graph the parabola.

To find the vertex, I look at the coefficients a, b, and c. The formula for the vertex gives me:

h = –b/2a = –(1)/2(3) = –1/6

Then I can find k by evaluating y at h = –1/6:

k = 3( –1/6 )2 + ( –1/6 ) – 2

= 3/36 – 1/6 – 2

= 1/12 – 2/12 – 24/12

= –25/12

So now I know that the vertex is at ( –1/6 , –25/12 ). Using the formula was helpful, because

this point is not one that I was likely to get on my T-chart.

I need additional points for my graph:

Now I can do my graph, and

The Best IIT-JEEPrepCourse by VidyamandirClasses Exclusive Priceon WizIQ. Save 10%www.wiziq.com/vidyamandir

IIT JEE PhysicsMaterialCheck Demo, 600 HrsVideo TutorialsPrepared by Topper.Try B4 you BuyExponentEducation.com/I…

CBSE Class VI to X3000+ solvedquestions & testsStudy from Home.Register freewww.LearnNext.com

Technical AnalysisWebnarLearn how to usecharting and technicalanalysis to trade.www.equis.com

IIT, AIEEE Tricks& TipsIIT & AIEEE 2012-13preparation. StudyMaterial-Test Series.Buy Nowvriti.com/Brilliant-Tutorials

I will label the vertex:

When you write down the vertex in your homework, write down the exact coordinates: "( –1/6 , –25/12 )".

But for graphing purposes, the decimal approximation of "(–0.2, –2.1)" may be more helpful, since it's

easier to locate on the axes.

The only other consideration regarding the vertex is the "axis of symmetry". If you look at a parabola,you'll notice that you could draw a vertical line right up through the middle which would split the parabolainto two mirrored halves. This vertical line, right through the vertex, is called the axis of symmetry. If

you're asked for the axis, write down the line "x = h", where h is just the x-coordinate of the vertex. So

in the example above, then the axis would be the vertical line x = h = –1/6.

Helpful note: If your quadratic's x-intercepts happen to be nice neat numbers (so they're relatively easy

to work with), a shortcut for finding the axis of symmetry is to note that this vertical line is always

exactly between the two x-intercepts. So you can just average the two intercepts to get the location of

the axis of symmetry and the x-coordinate of the vertex. However, if you have messy x-intercepts (as in

the example above) or if the quadratic doesn't actually cross the x-axis (as you'll see on the next page),

then you'll need to use the formula to find the vertex.

<< Previous Top | 1 | 2 | 3 | 4 | Return to Index Next >>

Cite this article as: Stapel, Elizabeth. "Graphing Quadratic Functions: The Leading Coefficient / The Vertex."

Purplemath. Available from http://www.purplemath.com/modules/grphquad2.htm.

Accessed 22 August 2011

Copyright © 2002-2011 Elizabeth Stapel | About | Terms of Use Feedback | Error?

Ads by Google Graphing Calculator Calculator Quadratic Live Graph Quadratic Formula Graphing Softw are

The Purplemath ForumsHelping students gain understanding

and self-confidence in algebra

Search

powered by FreeFind


Graphing Quadratic Functions: Examples (page 3 of 4)


Find the vertex and intercepts of y = 3x2 + x – 2 and graph; remember to label the

vertex and the axis of symmetry.

This is the same quadratic as in the last example. I already found the vertex when I worked the

problem above. This time, I also need to find the intercepts before I do my graph. To find the y-

intercept, I set x equal to zero and solve:

y = 3(0)2 + (0) – 2 = 0 + 0 – 2 = –2

Then the y-intercept is the point (0, –2). To find the x-intercept, I set y equal to zero, and solve:

0 = 3x2 + x – 2 0 = (3x – 2)(x + 1) 3x – 2 = 0 or x + 1 = 0 x = 2/3 or x = – 1

Then the x-intercepts are at the points (–1, 0) and ( 2/3, 0).

The axis of symmetry is halfway between the two x-

intercepts at (–1, 0) and at ( 2/3 , 0); using this, I can

confirm the answer from the previous page:

(–1 + 2/3) / 2 = (–1/3) / 2 = –1/6

The complete answer is a listing of the vertex, the axisof symmetry, and all three intercepts, along with a niceneat graph:

The vertex is at ( –1/6 , –25/12 ), the axis of

symmetry is the line x = –1/6 , and the intercepts

are at (0, –2), (–1, 0), and ( 2/3, 0).

Find the intercepts, the axis of symmetry, and vertex of y = x2 – x – 12.

To find the y-intercept, I set x equal to 0 and solve:

y = (0)2 – (0) – 12 = 0 – 0 – 12 = –12

To find the x-intercept, I set y equal to 0 and solve:

0 = x2 – x – 12

0 = (x – 4)(x + 3)

x = 4 or x = –3

To find the vertex, I look at the coefficients: a = 1 and b = –1. Plugging into the formula, I get:

h = –(–1)/2(1) = 1/2 = 0.5

Purplemath:


Printing pages

Donating

School l icensing

Reviews of

Internet Sites:

Free Help

Practice

Et Cetera

The "Homework

Guidelines"


Tutoring ($$)


your personal use.


CBSE Class VI to X3000+ solvedquestions & testsStudy from Home.Register freewww.LearnNext.com

preparing forIIT,AIEEE?Arihant Books + OnlineTest Series. 6000+Questions. Get OfferNow!vriti.com/Arihant-IIT-AIEEE

Preparing For CAT2011?Attend Interactive,Intensive & Live CrashCourse, Only Rs.5000www.WiZiQ.com/course/1…

Algebra EquationGraphingPlot and graphequations online; lines,circles, parabolas,cubics.www.WebGraphing.com

To find k, I plug h = 1/2 in for x inside y = x2 – x – 12, and simplify:

k = (1/2)2 – (1/2) – 12 = 1/4 – 1/2 – 12 = –12.25

Once I have the vertex, it's easy to write down the axis of symmetry: x = 0.5. Now I'll find some

additional plot points, to fill in the graph:

For my own convenience, I picked x-values that were centered around the x-coordinate of the

vertex. Now I can plot the parabola: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

The vertex is at the point (0.5, –12.25),

the axis of symmetry is the line x = 0.5,

and the intercepts are at the points (0, –12), (–3, 0), and (4, 0).

<< Previous Top | 1 | 2 | 3 | 4 | Return to Index Next >>

Cite this article as: Stapel, Elizabeth. "Graphing Quadratic Functions: Examples." Purplemath. Available from

http://www.purplemath.com/modules/grphquad3.htm. Accessed 22 August 2011


Ads by Google

Create Graph

Live Graph

Excel Graph

Quadratic Formula

Ads by Google Calculator Quadratic Multiple Axis Graph Quadratic Equation Plot a Graph Graph Paper

77 people onlineThey can help you right now!

Search

powered by FreeFind


Graphing Quadratic Functions: Examples (page 4 of 4)


Find the x-intercepts and vertex of y = –x2 – 4x + 2.

Since it is so simple to find the y-intercept (and it will probably be a point in my T-chart anyway),

they are only asking for the x-intercepts this time. To find the x-intercept, I set y equal 0 and

solve:

0 = –x2 – 4x + 2

x2 + 4x – 2 = 0

For graphing purposes, the intercepts are at about (–4.4, 0) and (0.4, 0). (When I write down

the answer, I will of course use the "exact" form, with the square roots; my calculator's decimalapproximations are just for helping me graph.)

To find the vertex, I look at the coefficients: a = –1 and b = –4. Then:

h = –(–4)/2(–1) = –2

To find k, I plug h = –2 in for x in y = –x2 – 4x + 2, and simplify:

k = –(–2)2 – 4(–2) + 2 = –4 + 8 + 2 = 10 – 4 = 6

Now I'll find some additional plot points, to help me fill in my graph:

Note that I picked x-values that were centered around the x-coordinate of the vertex. Now I'll plot

the parabola:

Purplemath:


Printing pages

Donating

School l icensing

Reviews of

Internet Sites:

Free Help

Practice

Et Cetera

The "Homework

Guidelines"


Tutoring ($$)


your personal use.

CBSE Class VI to X CBSE Maths Class VI to X Study from Home.No Internet Needed! www.LearnNext.com

The Best IIT-JEE Prep Course by Vidyamandir Classes Exclusive Price on WizIQ. Save 10% www.wiziq.com/vidyamandir

IIT JEE Physics Material Check Demo, 600 Hrs Video Tutorials Prepared by Topper. Try B4 you Buy ExponentEducation.com/IIT_Physics

The vertex is at (–2, 6), and the intercepts are at the following points:

(0, 2), , and

Find the x-intercepts and vertex of y = –x2 + 2x – 4.

To find the vertex, I look at the coefficients: a = –1 and b = 2. Then:

h = –(2)/2(–1) = 1

To find k, I'll plug h in for x and simplify:

k = –(1)2 + 2(1) – 4 = –1 + 2 – 4 = 2 – 5 = –3

The vertex is below the x-axis, and, since this is a negative quadratic, I know that the parabola is

going to be upside-down. So can my line possibly cross the x-axis? Can there possibly be any

x-intercepts? Of course not! So I expect to get "no (real) solution" when I try to find the x-

intercepts, but I need to show my work anyway. To find the x-intercept, I set y equal 0 and solve:

0 = –x2 + 2x – 4

x2 – 2x + 4 = 0

As soon as I get a negative inside the square root, I know that I can't get a graphable solution.

So, as expected, there are no x-intercepts. Now I'll find some additional plot points, to fill in my

graph:

Note that I picked x-values that were centered around the x-coordinate of the vertex. Now I'll plot

the parabola: Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved


The Best IIT-JEEPrepCourse by VidyamandirClasses Exclusive Priceon WizIQ. Save 10%www.wiziq.com/vidyamandir

CBSE Class VI to XCBSE Maths Class VIto X Study fromHome.No InternetNeeded!www.LearnNext.com

IIT, AIEEE Tricks& TipsIIT & AIEEE 2012-13preparation. StudyMaterial-Test Series.Buy Nowvriti.com/Brilliant-Tutorials

Technical AnalysisWebnarLearn how to usecharting and technicalanalysis to trade.www.equis.com

The vertex is at (1, –3), and the only intercept is at (0, –4).

This last exercise illustrates one way you can cut down a bit on your work. If you solve for the vertex

first, then you can easily tell if you need to continue on and look for the x-intercepts, or if you can go

straight on to plotting some points and drawing the graph. If the vertex is below the x-axis (that is, if the

y-value is negative) and the quadratic is negative (so the parabola opens downward), then there will be

no x-intercepts. Similarly, if the vertex is above the x-axis (that is, if the y-value is positive) and the

quadratic is positive (so the parabola opens upward), then there will be no x-intercepts.

In most of the graphs that I did (though not the first one), it just so happened that the points on the T-chart were symmetric about the vertex; that is, that the points "matched" on either side of the vertex.While a parabola is always symmetric about the vertical line through the vertex (the parabola's "axis"),

the T-chart points might not be symmetric. In particular, the T-chart points will not "match" if the x-

coordinate of the vertex is something other than a whole number or a half-number (such as "3.5").

Warning: Don't expect the plot-points always to "match up" on either side of the vertex; in particular,don't do half the points on your T-chart and then "fill in" the rest of your T-chart by assuming asymmetry that might not exist.

Other tips for graphing: If the parabola is going to be "skinny", then expect that you will get some verylarge values in your T-chart. You will either end up with a really tall graph or else a rather short T-chart. Ifthe parabola is going to be "fat", then expect that you will probably have to plot points with fractions ascoordinates. In either case, when you go to connect the dots to draw the parabola, you might find ithelpful to turn the paper sideways and first draw the really curvy part through the vertex, making surethat it looks nice and round. Then turn the paper back right-side-up and draw the "sides" of the parabola.

Warning: Draw your graphs big enough to be clearly seen by your instructor. If you're fitting more thantwo or maybe three graphs on one side of a standard sheet of paper, then you're drawing your graphsway too small.

<< Previous Top | 1 | 2 | 3 | 4 | Return to Index

Cite this article as: Stapel, Elizabeth. "Graphing Quadratic Functions: Examples." Purplemath. Available from

http://www.purplemath.com/modules/grphquad4.htm. Accessed 22 August 2011


Ads by Google Calculator Quadratic Live Graph Quadratic Formula Graphing Multiple Axis Graph

Date post:	06-Mar-2015
Category:	Documents
Upload:	amitakshay
View:	649 times
Download:	14 times

Quadratic Equation Compiled

Documents