Solving Quadratic Equations. Review of Solving Quadratic Equations ax 2 +bx +c = 0 A quadratic...

Solving Quadratic Equations

Review of Solving Quadratic Equations

ax2 +bx +c = 0A quadratic equation is a 2nd degree equation whose graph is a parabola.

€

(x − 2)2 = 49Solve with Perfect Square Binomial

€

(x − 2)2 = 49

x − 2 = ±7

€

x − 2 = +7

x = 9

€

x − 2 = −7

x = −5

1. Get Perfect Squares on Both Sides of Equation.

2. Take Square Root of Perfect Squares

3. Solve Positive Number & Neg. Number

4. Check Answers by Putting into Original Eq.

€

(x − 2)2 = 50Solve with Perfect Square Binomial

€

(x − 2)2 = 50

x − 2 = ± 25 2 = ±5 2

x = 2 ± 5 2

€

x = 2 + 5 2

1. Get Perfect Squares Binomial.

2. Take Square Root of Both Sides of Eq

3. Simplify4. Check Answers by

Putting into Original Eq.

€

x = 2 − 5 2

€

x 2 + 5x = 6

x 2 + 5x − 6 = 0

(x + 6)(x −1) = 0

x + 6 = 0 & x −1= 0

x = −6 x = +1

Solve Quadratic Eq. By Factoring

1. Set Eq = 0 2. Factor 3. Set Each Factor = 04. Check Answers by


€

x 2 + 5x = 6

What If You Can’t Factor The Equation?

€

x 2 + 4x = 36

x 2 + 4x − 36 = 0The “X” factor procedure doesn’t work which means itcan not be factored

-36

4

You can solve this equation by Completing The Square

Review of a Trinomial Square

(x+5)2 = x2 + 10x + 25

When you square a binomial you get a TRINOMIAL SQUARE

What is Completing The Square?

x2 + 10x +

WHAT WILL MAKE A TRINOMIAL SQUARE?

?

Now we will use these Algebra tiles to show Completing the Square

1

x x2

-1

-x -x2

x2 + 10x + ?

It takes 25 ones to complete this squareNote that 25 is half of b squared(10/2)2 = 52 = 25

b

2

2

Remember this to Complete The Square

b

2

2

Finding Half of b Squared

€

x 2 + 6x 6

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

= 32 = 9

b

Add to Complete the Square

€

x 2 + 6x 6

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

= 32 = 9

x 2 + 6x + 9

b

This is now a trinomial square

x2+6x+9 is a Trinomial Square

x2+6x+9x+3

x+3

(x+3)2 =x2 +6x+9

More Examples of

€

x 2 + 8x 8

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

= 42 =16

x 2 + 8x +16

€

x 2 − 8x -8

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

= −4( )2

=16

x 2 − 8x +16

Half of b Squared

Half of b Squared

More Examples of

€

x 2 + 3x 3

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

=3

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

=9

4

x 2 + 3x +9

4

€

x 2 + 3x +9

4= x +

3

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

Half of b Squared

More Examples of

€

x 2 + 7x 7

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

=7

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

=49

4

x 2 + 3x +49

4

€

x 2 + 3x +49

4= x +

7

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

Half of b Squared

Complete The Squareand State The Binomial Square

€

1. x 2 +10x

€

2. x 2 + 4x

€

3. x 2 + 20x

€

4. x 2 +11x

€

1. x 2 +10x +25 =

€

2. x 2 + 4x + 4 =

€

3. x 2 + 20x +100 =

€

4. x 2 +11x +121

4=


€

1. x 2 +10x +25 = (x +5)2

€

2. x 2 + 4x + 4 = (x +2)2

€

3. x 2 + 20x +100 = (x +10)2

€

4. x 2 +11x +121

4= x +

11

2

⎛

⎝ ⎜

⎞

⎠ ⎟2


Solve This Quad. Eq.

€

x 2 + 4x − 31= 0

If the “X” factor doesn’t work, you can’t solve by Factoring

-31

4

€

x 2 + 4x − 31= 0

Solve by Completing The Square

€

x 2 + 4x − 31= 0


€

x 2 + 4x = 31 1. First Put c on One Side of equation.

€

x 2 + 4x − 31 = 0


€

x 2 + 4x = 31 1. First Put c on One Side of equation.

2. Divide by a

Skip this step when a=1

€

x 2 + 4x − 31= 0


1. First Put c on One Side of equation.

2. Divide by a3. Complete The Square

ADD HALF OF b SQUARED

(b/2)2=(4/2)2=4

€

x 2 + 4x − 31= 0

€

x 2 + 4x = 31

x 2 + 4x + 4 = 31+ 4

b



2. Divide by a3. Complete The Square 4. Write Trinomial Square as a

Binomial Square

€

x 2 + 4x − 31= 0

€

x 2 + 4x = 31

x 2 + 4x + 4 = 31+ 4

(x + 2)2 = 35




Binomial Square5. Square Root both sides

€

x 2 + 4x − 31= 0

€

x 2 + 4x = 31

x 2 + 4x + 4 = 31+ 4

(x + 2)2 = 35

x +2 = ± 35




Binomial Square5. Square Root both sides6. Solve for x (Note 2 Answers)

€

x 2 + 4x − 31= 0

€

x 2 + 4x = 31

x 2 + 4x + 4 = 31+ 4

(x + 2)2 = 35

x +2 = ± 35

x = −2 ± 35

−2 + 35 & - 2 - 35



2. Divide equation by a3. Complete The Square

(b/2)2=

4. Write Trinomial Square as a Binomial Square

5. Square Root both sides6. Solve for x (Note 2 Answers)

€

x 2 + 8x − 3 = 0




(b/

2)2=(8/2)2=42=16



€

x 2 + 8x − 3 = 0

€

x 2 + 8x − 3 = 0

x 2 + 8x = 3

x 2 + 8x +16 = 3 +16

(x + 4)2 = 19

x + 4 = 19

x = −4 ± 19




(b/2)2=



€

3x 2 + 6x − 3 = 6




(b/2)2=(2/2)2=1



€

3x 2 + 6x = 9

x 2 + 2x + __ = 3+ __

x 2 + 2x +1 = 3 +1

(x +1)2 = 4

x +1= ± 4

x = −1± 2

x =1 or x = -3

€

3x 2 + 6x − 3 = 6


€

1. x 2 + 6x − 21 = 0

€

2. x 2 + 9x −1= 0

€

3. 2x 2 + 8x −11 = 0




(b/2)2= (6/2)2=32=94. Write Trinomial Square as a


€

x 2 + 6x − 21 = 0

x 2 + 6x = 21

x 2 + 6x +9 = 21+ 9

(x + 3)2 = 30

(x + 3)2 = 30

x + 3 = ± 30

x = −3± 30

€

1. x 2 + 6x − 21= 0




(b/2)2= (9/2)2= 81/44. Write Trinomial Square as a


€

x 2 + 9x −1 = 0

x 2 + 9x =1

x 2 + 9x + 81

4=1+

81

4

x +9

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

=85

4

x +9

2

⎛

⎝ ⎜

⎞

⎠ ⎟2

=± 85

4

x +9

2=

± 85

2

x = −9

2

± 85

2=

−9 ± 85

2

€

2. x 2 + 9x −1 = 0




(b/2)2= (2/2)2=14. Write Trinomial Square as a


€

3. 4x 2 + 8x −1 = 0

€

4x 2 + 8x =1

4x 2

4+

8x

4=

1

4

x 2 + 2x +1 =1

4+1 =

1

4+

4

4

x + 2( )2

=5

4

x + 2( )2

=± 5

4

x + 2 =± 5

2

x = −2 ± 5

2=

−1± 5

2

Solve Quadratic Equations With The Quadratic Formula

So far we have solved quadratic equations with FACTORING (Solves Equations with Integer Solutions) & COMPLETING THE SQUARE (Solves All quadratic equations)

THE QUADRATIC FORMULA ALSO SOLVES ALL QUADRATIC EQUATIONS

QUADRATIC EQUATION

€

x =−b ± b2 − 4ac

2a

€

- b

2 a

€

b2 − 4ac

+

2

X=

Solve with Quadratic Equation

€

x 2 + 4x = −4


1. Put Equation into Std. Form (ax2+bx+c=0)

€

x 2 + 4x = −4

€

x 2 + 4x + 4 = 0a=1 b c



2. Plug a, b & c into the Quad. Eq.

€

x 2 + 4x = −4

€

x =−b ± b2 − 4ac

2a

€

x 2 + 4x + 4 = 0

x =−4 ± 42 − 4(4)

2

a=1 b c

€

x =−b ± b2 − 4ac

2a




3. Simplify Radical

€

x 2 + 4x = −4

€

x 2 + 4x + 4 = 0

x =−4 ± 42 − 4(4)

2

x =−4 ± 0

2=

−4

2

a=1 b c

€

x =−b ± b2 − 4ac

2a




3. Simplify Radical4. Solve for x

€

x 2 + 4x = −4

€

x 2 + 4x + 4 = 0

x =−4 ± 42 − 4(4)

2

x =−4 ± 0

2=

−4

2x = −2

a=1 b c


€

x 2 + 2x = 2


€

x 2 + 2x − 2 = 01. Put Equation into Std. Form

(ax2+bx+c=0)

€

x 2 + 2x = 2

a=1 b c

€

x =−b ± b2 − 4ac

2a


€

x 2 + 2x − 2 = 0

x =−2 ± 4 − 4(−2)

2



€

x 2 + 2x = 2

a=1 b c

€

x =−b ± b2 − 4ac

2a


€

x 2 + 2x − 2 = 0

x =−2 ± 4 − 4(−2)

2

x =−2 ± 12

2=

−2 ± 4 3

2

x =−2 ± 2 3

2



3. Simplify Radical

€

x 2 + 2x = 2

a=1 b c

€

x =−b ± b2 − 4ac

2a


€

x 2 + 2x − 2 = 0

x =−2 ± 4 − 4(−2)

2

x =−2 ± 12

2=

−2 ± 4 3

2

x =−2 ± 2 3

2= −1± 3




€

x 2 + 2x = 2

a=1 b c

Solve by quadratic equation

€

1. x 2 + 2x − 7 = 0

€

2. x 2 + 5x = −3

€

3. 3x 2 − 7 = 4x

€

x =−b ± b2 − 4ac

2a





a=1 b c

€

1. x 2 + 2x − 7 = 0

€

x 2 + 2x − 7 = 0

x =−2 ± 2( )

2− 4 1⋅−7( )

2 ⋅1

x =−2 ± 4 + 28

2=

−2 ± 32

2

x =−2 ± 16 2

2=

−2 ± 4 2

2

x = −1± 2 2

€

x =−b ± b2 − 4ac

2a





a=1 b c

€

2. x 2 + 5x = −3

€

x 2 + 5x + 3 = 0

x =−5 ± 25 − 4(3)

2

x =−5 ± 13

2

€

x =−b ± b2 − 4ac

2a





a b c

€

3x 2 + 4x − 7 = 0

x =−4 ± 14 − 4(−21)

3⋅2

x =−4 ± 85

6

€

3. 3x 2 − 7 = 4x

The DiscriminantThe Discriminant equals b2-4ac

€

x =−b ± b2 − 4ac

2a

The Discriminant is what’s under the RADICAL


€

x =−b ± b2 − 4ac

2a2 Solutions if positive1 Solution if 0 0 Solutions if negative

The Discriminant tells you how many solutions or roots

x

•

•

•

•

This quadratic has two roots so the Discriminant must be positive

y


This quadratic has one roots so the Discriminant must be zero

X


This quadratic has no roots so the Discriminant must be negative

X

Find The Number of Solutions (Roots)

€

3x 2 + 4x + 2 = 0

b2 − 4(ac)

42 − 4(3⋅2)

16 − 24 = −8

€

3x 2 + 4x + 2 = 0

1. Identify a, b & c.2. Put values into the

discriminant b2-4ac3. If Pos. 2 Solutions

If Zero 1 Solution If Neg. 0 SolutionsThe discriminant is neg. so

there are no solutions

Determine The Number of Solutions or Roots

€

1. x 2 − x + 2 = 0

€

2. 2x 2 − 3x = 8

€

3. 2x 2 − 4x − 7 = −15

Determine The Number of Solutions or Roots

€

1. x 2 − x + 2 = 0 a =1, b = -7, c = 2

b2 − 4(ac) = (−1)2 − 4(1⋅2) =1− 8 = −7

€

2. 2x 2 − 3x = 8

2x 2 − 3x − 8 = 0 a = 2, b = -3, c = -8

b2 − 4(ac) = (−3)2 − 4(2 ⋅−8)

= 9 − 4(−16) = 9 + 56 = 65

€

3. 2x 2 − 4x − 7 = −15

2x 2 − 4x + 8 = 0 a = 2, b = -4, c = 8

b2 − 4(ac) = (−4)2 − 4(2 ⋅8) =16 −16 = 0

Singing the Quadratic Equation(To the melody of Pop Goes The Weasel)

Minus B plus or minus the square rootB squared minus 4ACAll over 2A, That’s the quadratic equation.

It’s the greatest equation of all time.It even has a wonderful rhyme.If you don’t know it, It would be a crimeThat’s the quadratic equation.

Math is the most taught subject in the worldSo you’d better believe meBone up to pass the SATYou won’t end up a weasel.

SOLVINGQUADRATICEQUATIONS

€

( x − 2 )

2

= 49

Solve with PerfectSquare Binomial

€

( x − 2 )

2

= 49

x − 2 = ± 7

€

x − 2 = + 7

x = 9

€

x − 2 = − 7

x = − 5

1. Get Perfect Squares onBoth Sides of Equation.

2. Take Square Root ofPerfect Squares

3. Solve Positive Number& Neg. Number

4. Check Answers byPutting into Original Eq.

Linear Vs. QuadraticStraight Line

First Degree Eq.

Exponent is 1

ax +by +c = 0

y = mx + b

Parabola or “U” Shaped

Second Degree Equation

Exponent is 2

f(x) = ax2 +bx +c = 0


€

x =

− b ± b

2

− 4 ac

2 a

2 Solutions if positive1 0 Solution if0 Solutions ifnetative

X

•

•

•

•€

x =− b ± b

2− 4 ac

2 a

Solve withQuadratic Equation

€

x2

+ 2 x − 2 = 0

x =− 2 ± 4 − 4 ( − 2 )

2

x =

− 2 ± 12

2

=

− 2 ± 4 3

2

x =− 2 ± 2 3

2

= − 1 ± 3

1. Put Equation into Std. Form(ax2+bx+c=0)



€

x2

+ 2 x = 2

a=1 b c

Solve by CompletingThe Square

1. First Put c on Right Hand Sideof equation.


(b/2)2=(2/2)2=14. Write Trinomial Square as a


€

3 x2

+ 6 x = 9

x2

+ 2 x + __ = 3 + __

x2

+ 2 x + 1 = 3 + 1

( x + 1 )2

= 4

x + 1 = ± 4

x = − 1 ± 2

x = 1 or x = -3

€

3 x2

+ 6 x − 3 = 6

€

x2

+ 5 x = 6

x2

+ 5 x − 6 = 0

( x + 6 )( x − 1 ) = 0

x + 6 = 0 & x − 1 = 0

x = − 6 x = + 1

Solve Quadratic Eq.By Factoring

1. Set Eq = 02. Factor3. Set Each Factor = 04. Check Answers by


€

x2

+ 5 x = 6

What Method To Use

• SquareRoot when there is a perfect square on one side of eq.• Use “X” Factor when trinomial is factorable• Use quadratic formula or complete the square if not factorable

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Solving Quadratic Equations. Review of Solving Quadratic Equations ax 2 +bx +c = 0 A quadratic...

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