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Solving Quadratic Equations Using the Quadratic Formula MA.912.A.7.2 Solve quadratic equations over the real numbers by factoring and by using the quadratic formula.
Solving Quadratic Equations Using the Quadratic Formula
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
2a
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
a=
0
a
x2 +b
ax +
c
a= 0
x2 +b
ax = −
c
a
€
= Divide both sides by “a”.
Subtract constant fromboth sides.
Deriving the Quadratic Formula by Completing the Square.
€
x2 +b
ax + = −
c
a+
x +b
2a
⎛ ⎝ ⎜
⎞ ⎠ ⎟2
= −c
a+b2
4a2
x +b
2a
⎛ ⎝ ⎜
⎞ ⎠ ⎟2
=b2 − 4ac
4a2
Complete The Square
€
b
2a
⎛ ⎝ ⎜
⎞ ⎠ ⎟2
€
b
2a
⎛ ⎝ ⎜
⎞ ⎠ ⎟2
Factor thePerfect SquareTrinomial
Simplify expressionon the left side byfinding the LCD
Deriving the Quadratic Formula by Completing the Square.
€
x +b
2a
⎛ ⎝ ⎜
⎞ ⎠ ⎟2
=b2 − 4ac
4a2
x +b
2a=
b2 − 4ac
4a2
x +b
2a= ±
b2 − 4ac
2a
Take the squareroot of both sides
€
Don't Forget :
x2 = x
Solve absolute value/Simplify radical
Deriving the Quadratic Formula by Completing the Square.
€
x +b
2a= ±
b2 − 4ac
2a
x = −b
2a±
b2 − 4ac
2a
x =−b ± b2 − 4ac
2a
Isolate x
Simplify
Congratulations!You have derivedThe Quadratic Formula
#1 Solve using the quadratic formula.
€
3x 2 − 7x + 2 = 0
a
acbbx
2
42
2 ,7 ,3 cba
)3(2
)2)(3(4)7()7( 2 x
6
24497 x
6
57x
6
12x
6
2x
€
x = 2 or x =1
3
6
257x
Graph
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
€
x =1
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!
#2 Solve using the quadratic formula.
€
2x 2 − 4x = 5
a
acbbx
2
42
5 ,4 ,2 cba
)2(2
)5)(2(4)4()4( 2 x
4
40164 x
4
564x
4
1424x
0542 2 xx
€
x ≈ 2.87 or x ≈ − 0.87€
x =2 ± 14
2
€
x ≈2 ± 3.74
2
ExactSolution
ApproxSolution
Graph
Clink on link for graphing calculator.http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html
€
y = 2x 2 − 4x − 5
#3 Solve using the quadratic formula
€
x 2 = − 5x − 7
a
acbbx
2
42
€
a =1, b = 5, c = 7
)1(2
)7)(1(455 2 x
2
28255 x
€
x =−5 ± −3
2
€
x 2 + 5x + 7 = 0
€
−3The is not a real number, therefore this equation has ‘NO Real Solution’
Graph
Clink on link for graphing calculator.http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html
€
y = x 2 + 5x + 7
#4 Solve using the quadratic formula
xx 16642
a
acbbx
2
42
64 ,16 ,1 cba
)1(2
)64)(1(41616 2 x
2
25625616 x
2
016x
€
x = 8
064162 xx
Would factoring work to solve this equation?
€
x − 8( ) x − 8( ) = 0
Graph
Clink on link for graphing calculator.http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html
€
y = x 2 −16 + 64
#5 Solve using the quadratic formula.
€
2x +1 = x 2
a
acbbx
2
42
€
a = −1, b = 2, c =1
€
x =−(2) ± (2)2 − 4(−1)(1)
2(−1)
€
x =−2 ± 4 + 4
−2
€
x =−2 ± 8
−2
€
x =−2 ± 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
acbbx
2
42
€
a =1, b = −2, c = −1
€
x =−(−2) ± (−2)2 − 4(1)(−1)
2(1)
€
x =2 ± 4 + 4
2
€
x =2 ± 8
2
€
x =2 ± 2 2
2
€
x 2 − 2x −1 = 0
€
x ≈ 2.41 or x ≈ − 0.41€
x =1 ± 2
€
x ≈1±1.41
ExactSolution
ApproxSolution
Graph
Clink on link for graphing calculator.http://my.hrw.com/math06_07/nsmedia/tools/Graph_Calculator/graphCalc.html
€
y = −x 2 + 2x +1 & y = x 2 − 2x −1
The Discriminant
€
x =−b ± b2 − 4ac
2a
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
€
= 25
2 RealRationalSolutions
€
4x 2 + 5 = x
acb 4 nt discrimina 2 5 ,1 ,4 cba
€
= (−1)2 − 4(4)(5)
€
= 1− 80
€
= − 79
#7 Find the discriminant and describe the solutions to the equations.
€
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
€
=56
#8 Find the discriminant and describe the solutions to the equations.
€
a = 2, b = −4, c = −5
2 RealIrrationalSolutions
484 2 xx
acb 4 nt discrimina 2
€
= (−8)2 − 4(4)(4)
€
= 64 − 64
€
= 0
#9 Find the discriminant and describe the solutions to the equations.
€
4x 2 − 8x + 4 = 0
€
a = 4, b = −8, c = 4
1 RealRationalSolution
