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QUADRATIC EQUATIONS
Math Center Workshop SeriesJanice Levasseur
Basics
• A quadratic equation is an equation equivalent to an equation of the type
ax2 + bx + c = 0, where a is nonzero
• We can solve a quadratic equation by factoring and using The Principle of Zero Products
If ab = 0, then either a = 0, b = 0, or both a and b = 0.
Ex: Solve (4t + 1)(3t – 5) = 0
Notice the equation as given is of the form ab = 0
set each factor equal to 0 and solve
4t + 1 = 0 Subtract 1
4t = – 1 Divide by 4
t = – ¼
3t – 5 = 0 Add 5
3t = 5 Divide by 3
t = 5/3
Solution: t = - ¼ and 5/3 t = {- ¼, 5/3}
Ex: Solve x2 + 7x + 6 = 0
Quadratic equation factor the left hand side (LHS)
x2 + 7x + 6 = (x + )(x + )6 1
x2 + 7x + 6 = (x + 6)(x + 1) = 0
Now the equation as given is of the form ab = 0
set each factor equal to 0 and solve
x + 6 = 0x = – 6
x + 1 = 0x = – 1
Solution: x = - 6 and – 1 x = {-6, -1}
Ex: Solve x2 + 10x = – 25
Quadratic equation but not of the form ax2 + bx + c = 0
x2 + 10x + 25 = (x + )(x + )5 5
x2 + 10x + 25 = (x + 5)(x + 5) = 0
Now the equation as given is of the form ab = 0
set each factor equal to 0 and solve
x + 5 = 0
x = – 5
x + 5 = 0
x = – 5
Solution: x = - 5 x = {- 5} repeated root
Quadratic equation factor the left hand side (LHS)
Add 25 x2 + 10x + 25 = 0
Ex: Solve 12y2 – 5y = 2
Quadratic equation but not of the form ax2 + bx + c = 0
ac method a = 12 and c = – 2
= (3y – 2)(4y + 1)
Quadratic equation factor the left hand side (LHS)
Subtract 2 12y2 – 5y – 2 = 0
ac = (12)(-2) = - 24 factors of – 24 that sum to - 5
1&-24, 2&-12, 3&-8, . . . 12y2 – 5y – 2 = 12y2 + 3y – 8y – 2
= 3y(4y + 1) – 2(4y + 1)
12y2 – 5y – 2 = 0
Now the equation as given is of the form ab = 0
set each factor equal to 0 and solve
3y – 2 = 0
3y = 2
4y + 1 = 0
4y = – 1
Solution: y = 2/3 and – ¼ y = {2/3, - ¼ }
y = 2/3 y = – ¼
12y2 – 5y – 2 = (3y - 2)(4y + 1) = 0
Ex: Solve 5x2 = 6x
Quadratic equation but not of the form ax2 + bx + c = 0
5x2 – 6x = x( )5x – 6
5x2 – 6x = x(5x – 6) = 0
Now the equation as given is of the form ab = 0
set each factor equal to 0 and solve
x = 05x – 6 = 05x = 6
Solution: x = 0 and 6/5 x = {0, 6/5}
Quadratic equation factor the left hand side (LHS)
Subtract 6x 5x2 – 6x = 0
x = 6/5
Completing the Square
• Recall from factoring that a Perfect-Square Trinomial is the square of a binomial:Perfect square Trinomial Binomial Square x2 + 8x + 16 (x + 4)2
x2 – 6x + 9 (x – 3)2
• The square of half of the coefficient of x equals the constant term: ( ½ * 8 )2 = 16 [½ (-6)]2 = 9
Completing the Square
• Write the equation in the form x2 + bx = c
• Add to each side of the equation [½(b)]2
• Factor the perfect-square trinomial x2 + bx + [½(b)] 2 = c + [½(b)]2
• Take the square root of both sides of the equation
• Solve for x
Ex: Solve w2 + 6w + 4 = 0 by completing the square
First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1.
w2 + 6w = – 4
Add [½(b)]2 to both sides: b =
6
6 [½(6)]2 = 32 = 9
w2 + 6w + 9 = – 4 + 9
w2 + 6w + 9 = 5
(w + 3)2 = 5
Now take the square root of both sides
5)3w( 2
53w
53w }53,53{w
Ex: Solve 2r2 = 3 – 5r by completing the square
First, rewrite the equation with the constant on one side of the equals and a lead coefficient of 1.
2r2 + 5r = 3
Add [½(b)]2 to both sides: b =
(5/2)
5/2 [½(5/2)]2 = (5/4)2 = 25/16
r2 + (5/2)r + 25/16 = (3/2) + 25/16
r2 + (5/2)r + 25/16 = 24/16 + 25/16
(r + 5/4)2 = 49/16
Now take the square root of both sides
r2 + (5/2)r = (3/2)
16/49)4/5r( 2
)4/7(4/5r )4/7()4/5(r
r = - (5/4) + (7/4) = 2/4 = ½
and r = - (5/4) - (7/4) = -12/4 = - 3
r = { ½ , - 3}
Ex: Solve 3p – 5 = (p – 1)(p – 2)
Is this a quadratic equation? FOIL the RHS
3p – 5 = p2 – 2p – p + 2
3p – 5 = p2 – 3p + 2
p2 – 6p + 7 = 0
Collect all terms
A-ha . . .
Quadratic Equation complete the square
p2 – 6p = – 7 [½(-6)]2 = (-3)2 = 9
p2 – 6p + 9 = – 7 + 9
(p – 3)2 = 2
2)3p( 2
23p 23p
}23,23{p
The Quadratic Formula
• Consider a quadratic equation of the form ax2 + bx + c = 0 for a nonzero
• Completing the square 2ax bx c
2b c
x xa a
2 2
2
2 2
b b c bx x
a 4a a 4a
The Quadratic Formula
Solutions to ax2 + bx + c = 0 for a nonzero are
22
2
b b 4acx2a 4a
2b b 4acx
2a
2 2
2
2 2 2
b b 4ac bx x
a 4a 4a 4a
Ex: Use the Quadratic Formula to solve x2 + 7x + 6 = 0
Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by
a2ac4bb
x2
Identify a, b, and c in ax2 + bx + c = 0:
a = b = c = 1
1
7
7
6
6
Now evaluate the quadratic formula at the identified values of a, b, and c
)1(2)6)(1(477
x2
224497
x
2257
x
257
x
x = ( - 7 + 5)/2 = - 1 and x = (-7 – 5)/2 = - 6
x = { - 1, - 6 }
Ex: Use the Quadratic Formula to solve
2m2 + m – 10 = 0Recall: For quadratic equation ax2 + bx + c = 0, the solutions to a quadratic equation are given by
a
acbbm
2
42
Identify a, b, and c in am2 + bm + c = 0:
a = b = c = 2
2
1
1
- 10
– 10
Now evaluate the quadratic formula at the identified values of a, b, and c
)2(2)10)(2(411
m2
48011
m
4811
m
491
m
m = ( - 1 + 9)/4 = 2 and m = (-1 – 9)/4 = - 5/2
m = { 2, - 5/2 }
Any questions . . .