Solving Quadratic Equations Using the Quadratic Formula

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MA.912.A.7.2 Solve quadratic equations over the real numbers by factoring and by using the quadratic

formula.

The Quadratic Formula

x =−b ± b2 − 4ac

The solutions of a quadratic equation written in Standard Form, ax2 + bx + c = 0, can be found by Using the Quadratic Formula.

Click on the link below to view a song to help you memorize it.

http://www.regentsprep.org/Regents/math/algtrig/ATE3/quadsongs.htm

Deriving the Quadratic Formula by Completing the Square.

ax2 + bx + c

ax = −

= Divide both sides by “a”.

Subtract constant fromboth sides.

ax + = −

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

= −c

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

=b2 − 4ac

Complete The Square

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

Factor thePerfect SquareTrinomial

Simplify expressionon the left side byfinding the LCD

⎛ ⎝ ⎜

⎞ ⎠ ⎟2

=b2 − 4ac

b2 − 4ac

2a= ±

b2 − 4ac

Take the squareroot of both sides

Don't Forget :

x2 = x

Solve absolute value/Simplify radical

2a= ±

b2 − 4ac

x = −b

b2 − 4ac

x =−b ± b2 − 4ac

Isolate x

Simplify

Congratulations!You have derivedThe Quadratic Formula

#1 Solve using the quadratic formula.

3x 2 − 7x + 2 = 0

2 ,7 ,3 cba

)2)(3(4)7()7( 2 x

24497 x

x = 2 or x =1

Clink on link for graphing calculator.http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html

y = 3x 2 − 7x + 2

#1 Solve by factoring

3x 2 − 7x + 2 = 0

3x −1( ) x − 2( ) = 0

3x −1( ) = 0 or x − 2( ) = 0

3or x = 2

#2 Solve by factoring

2x 2 − 4x = 50542 2 xx

2x 5( ) x 1( ) = 0

2x 1( ) x 5( ) = 0This quadratic is Prime (will not factor),The Quadratic Formula must be used!

2x 2 − 4x = 5

5 ,4 ,2 cba

)5)(2(4)4()4( 2 x

40164 x

0542 2 xx

x ≈ 2.87 or x ≈ − 0.87€

x =2 ± 14

x ≈2 ± 3.74

ExactSolution

ApproxSolution

y = 2x 2 − 4x − 5

#3 Solve using the quadratic formula

x 2 = − 5x − 7

a =1, b = 5, c = 7

)7)(1(455 2 x

28255 x

x =−5 ± −3

x 2 + 5x + 7 = 0

−3The is not a real number, therefore this equation has ‘NO Real Solution’

y = x 2 + 5x + 7

#4 Solve using the quadratic formula

xx 16642

64 ,16 ,1 cba

)64)(1(41616 2 x

25625616 x

064162 xx

Would factoring work to solve this equation?

x − 8( ) x − 8( ) = 0

y = x 2 −16 + 64

2x +1 = x 2

a = −1, b = 2, c =1

x =−(2) ± (2)2 − 4(−1)(1)

2(−1)

x =−2 ± 4 + 4

x =−2 ± 8

x =−2 ± 2 2

−x 2 + 2x +1 = 0

x ≈ 2.41 or x ≈ − 0.41€

x =1 ± 2

x ≈1±1.41

ExactSolution

ApproxSolution

#5 What if we move everything to the right side?

2x +1 = x 2

a =1, b = −2, c = −1

x =−(−2) ± (−2)2 − 4(1)(−1)

x =2 ± 4 + 4

x =2 ± 8

x =2 ± 2 2

x 2 − 2x −1 = 0

x ≈ 2.41 or x ≈ − 0.41€

x =1 ± 2

x ≈1±1.41

ExactSolution

ApproxSolution

y = −x 2 + 2x +1 & y = x 2 − 2x −1

The Discriminant

x =−b ± b2 − 4ac

The expression inside the radical in the quadratic formula is called the Discriminant.

The discriminant can be used to determine the number of solutions that a quadratic has.

Understanding the discriminantDiscriminant

acb 42 # of real solutions

042 acb

042 acb1 real rational

solution

042 acb No real solution

Perfect square

NotPerfect

2 real rational solutions

2 real irrational solutions

#6 Find the discriminant and describe the solutions to the equations.

0134 2 yy

acb 4 nt discrimina 2 1 ,3 ,4 cba

= (3)2 − 4(4)(−1)

= 9 +16

2 RealRationalSolutions

4x 2 + 5 = x

acb 4 nt discrimina 2 5 ,1 ,4 cba

= (−1)2 − 4(4)(5)

= 1− 80

= − 79

4x 2 − x + 5 = 0

No RealSolutions

542 2 xx

acb 4 nt discrimina 2

0542 2 xx

= (−4)2 − 4(2)(−5)

= 16 + 40

a = 2, b = −4, c = −5

2 RealIrrationalSolutions

484 2 xx

acb 4 nt discrimina 2

= (−8)2 − 4(4)(4)

= 64 − 64

4x 2 − 8x + 4 = 0

a = 4, b = −8, c = 4

1 RealRationalSolution

Solving Quadratic Equations Using the Quadratic Formula

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