Easy way of solving quadratic equations

Step 1: multiply the coefficient of first term with the last term.

Step 2: Find the 2 factors of 6 that have the product of 6 and the sum or difference of -5.

Step 3: Adjust signs to get -5 & 6.

Step 4: Divide by 2 the coefficient.

Step 5: Make the signs opposite.

So, the 2 roots are x= 3/2 and x=1

1. FOILWhen two binomials are multiplied, their product can be expanded by FOIL. The letters stand for First, Outer, Inner, Last.Example 1:  Expand (2x + 5)(3x + 4)Solution:We multiply 2x by each of the terms in 3x + 4 and then multiply 5 by each term in 3x + 4, and combine the like terms in the middle.  2x and 3x are the First terms of each binomial, 2x and 4 are the Outer terms, 5 and 3x are the Inner terms, and 5 and 4 are the Last terms.We get:  (2x + 5)(3x + 4) = 6x2 + 23x + 20

Example 2:  Expand (x – 3)(2x + 1) Think of this as:  (x + –3)(2x + 1)So (x – 3)(2x + 1) = 2x2 – 5x – 3

2. Reverse FOIL" is a method used for factoring a quadratic trinomial. The strategy is to determine the First and Last

terms of each binomial in the factored product so the Outer and Inner products add to the middle term.

Example:  Factor 3x2 + 5x – 12Solution:  First we concentrate on 3x2 and –12.  3x2 is the product of 3x and x, so we place these First terms of the factors:3x2 + 5x – 12 = (3x     )(x     )Now –12 could result in many different ways.  It could be –1 times 12, –2 times 6, 2 times –6, or any number of other products.To make it easier, we ignore the negative for the time being and try various products that give 12, and then ask ourselves if it is possible to get the middle term, 5x, by subtracting the Outer and Inner products:3x2 + 5x – 12 = (3x     2)(x      6)Outer = 18x, Inner = 2x  No3x2 + 5x – 12 = (3x     6)(x       2)Outer = 6x, Inner = 6x  No3x2 + 5x – 12 = (3x    4)(x      3)Outer = 9x, Inner = 4x  Yes, 9x –4x = 5xSo we want the 4x to be negative.  Therefore the factorization is:3x2 + 5x – 12 = (3x – 4)(x + 3)

3. Factoring a GCF firstSometimes a trinomial has a greatest common factor. When this is the case, we should first factor that, and then apply reverse FOIL.

Example:  Factor 6x2 – 2x – 4Solution: Every term is a multiple of 2.  That is, 2 is a greatest common factor, so we first factor that:6x2 – 2x – 4

= 2(3x2 – x – 2)Then we factor the trinomial in parentheses:6x2 – 2x – 4 = 2(3x2 – x – 2) = 2(3x + 2)(x – 1)

4. The Difference of Squares Some quadratic binomials are the difference of two perfect squares. When this is the case, they can be factored as a difference times a sum, and the following formula applies:a2 – b2 = (a – b)(a + b)

Example: Factor 9x2 – 4Solution: 9x2 = 3x.3x and 4 = 2.2, so this binomial is the difference of two squares. It factors as:9x2 – 4 = (3x – 2)(3x + 2)

Example 2: Factor 3x2 – 48Solution: 3x2 and 48 are not perfect squares, but both have 3 as a common factor. Therefore we start by factoring out 3:3x2 – 48 = 3(x2 – 16)Now x2 – 16 is the difference of squares, so it can be factored:3x2 – 48 = 3(x2 – 16) = 3(x – 4)(x + 4)

5. Solving Equations by FactoringWhen the product of two numbers is 0, then one of the numbers must be 0. This is a special property of 0 and does not apply to other numbers. For example, if the product of two numbers is 8, then neither number need be 8 (2 times 4 is 8, so the numbers could be 2 and 4).

This property of 0 is called the "zero product property" and can be used to solve an equation when its factored form is equal to 0.

For example, the equation (x – 5)(x + 3) = 0 tells us that the product of the two numbers x – 5 and x + 3 is 0. Therefore one of these numbers must be 0. That is, either x – 5 = 0 or x + 3 = 0. The solutions to these two equations are x = 5 and x = –3, so the equation (x – 5)(x + 3) = 0 has these two solutions.

Quadratic equations of the form ax2 + bx + c = 0 can often be solved by factoring the left side.Example: Solve x2 + 3 = 4xSolution: First subtract 4x from both sides to get: x2 – 4x + 3 = 0Then factor the left side: (x – 1)(x – 3) = 0Now set each factor to 0 and solve:x – 1 = 0 --> x = 1x – 3 = 0 --> x = 3

Type 2 QF

Finding the Discriminant

Using the formula with D calculated before

Vedic Quadratic formula = quick w derivative

Solve x(x – 2) = 4x(x – 2) = 4x2 – 2x = 4x2 – 2x – 4 = 0• Since there are no factors of (1)(–4) = –4 that add up

to –2, then this quadratic does not factor, use the Quadratic Formula; in this case, a = 1, b = –2, and c = –4:  

Then the answer is:  x = –1.24, x = 3.24, rounded to two places.

Quadratic Formula: Just plug in the values of a, b and c, and do the calculations. The ± means there are TWO answers.

Discriminant b2 - 4ac in the formula is called the Discriminant, because it can "discriminate" between the possible types of answer:when b2 - 4ac is positive, we get two Real solutions when it is zero we get just ONE real solution (both answers are the same) when it is negative we get two Complex solutions.

Using the Quadratic FormulaPut the values of a, b and c into the Quadratic Formula, and do the calculations.

Example: Solve 5x² + 6x + 1 = 0

Coefficients are: a = 5, b = 6, c = 1

Quadratic Formula: x = [ -b ± √(b2-4ac) ] / 2a

Put in a, b and c: x = [ -6 ± √(62-4×5×1) ] / (2×5)

Solve: x = [ -6 ± √(36-20) ]/10

x = [ -6 ± √(16) ]/10x = ( -6 ± 4 )/10x = -0.2 or -1

Answer: x = -0.2 or x = -1

Check -0.2: 5×(-0.2)² + 6×(-0.2) + 1= 5×(0.04) + 6×(-0.2) + 1= 0.2 -1.2 + 1= 0

Check -1: 5×(-1)² + 6×(-1) + 1= 5×(1) + 6×(-1) + 1= 5 - 6 + 1= 0

 

Complex Solutions: When the Discriminant (the value b2 - 4ac) is negative we get Complex solutions. It means our answer will include Imaginary Numbers.

Example: Solve 5x² + 2x + 1 = 0Coefficients are: a = 5, b = 2, c = 1

Note that The Discriminant is negative: b2 - 4ac = 22 - 4×5×1 = -16

Use the Quadratic Formula: x = [ -2 ± √(-16) ] / 10

The square root of -16 is 4i (i is √-1)

So: x = ( -2 ± 4i )/10Answer:

The graph does not cross the x-axis. That is why we ended up with complex numbers. In some ways it is easier: we don't need more

calculation, just leave it as

Summary

Quadratic Equation in Standard Form: ax2 + bx + c = 0

Quadratic Equations can be factored

Quadratic Formula: x = [ -b ± √(b2-4ac) ] / 2a

When the Discriminant (b2-4ac) is:

positive, there are 2 real solutions

zero, there is one real solution

negative, there are 2 complex solutions