QUADRATIC EQUATIONS

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QUADRATIC EQUATIONS
MSJC San Jacinto Campus Math Center Workshop
Series Janice Levasseur
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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.

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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
Divide by 4
4t 1
t ¼
Add 5
3t 5 0
3t 5
Divide by 3
t 5/3
Solution t - ¼ and 5/3 ? t - ¼, 5/3
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Ex Solve x2 7x 6 0
Quadratic equation ? factor the left hand side
(LHS)
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x2 7x 6 (x )(x )
? 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 0
x 1 0
x 1
x 6
Solution x - 6 and 1 ? x -6, -1
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Ex Solve x2 10x 25
Quadratic equation but not of the form ax2 bx
c 0
? x2 10x 25 0
Add 25
Quadratic equation ? factor the left hand side
(LHS)
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x2 10x 25 (x )(x )
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 0
x 5
x 5
Solution x - 5 ? x - 5 ? repeated root
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Ex Solve 12y2 5y 2
Quadratic equation but not of the form ax2 bx
c 0
? 12y2 5y 2 0
Subtract 2
Quadratic equation ? factor the left hand side
(LHS)
ac method ? a 12 and c 2
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)
(3y 2)(4y 1)
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? 12y2 5y 2 0
? 12y2 5y 2 (3y - 2)(4y 1) 0
Now the equation as given is of the form ab 0
? set each factor equal to 0 and solve
3y 2 0
4y 1 0
3y 2
4y 1
y 2/3
y ¼
Solution y 2/3 and ¼ ? y 2/3, - ¼
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Ex Solve 5x2 6x
Quadratic equation but not of the form ax2 bx
c 0
? 5x2 6x 0
Subtract 6x
Quadratic equation ? factor the left hand side
(LHS)
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5x2 6x x( )
5x
? 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
5x 6 0
x 0
5x 6
x 6/5
Solution x 0 and 6/5 ? x 0, 6/5
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Solving by taking square roots

• An alternate method of solving a quadratic
  equation is using the Principle of Taking the
  Square Root of Each Side of an Equation
• If x2 a, then

x
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Ex Solve by taking square roots 3x2 36 0
3x2 36 0
First, isolate x2
3x2 36
x2 12
Now take the square root of both sides
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Ex Solve by taking square roots 4(z 3)2 100
First, isolate the squared factor
4(z 3)2 100
(z 3)2 25
Now take the square root of both sides
z 3 5
z 3 5
? z 3 5 8 and z 3 5 2
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Ex Solve by taking square roots 5(x 5)2 75
0
First, isolate the squared factor
5(x 5)2 75
(x 5)2 15
Now take the square root of both sides
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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

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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

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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
6
Add ½(b)2 to both sides
b
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
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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
(5/2)
? r2 (5/2)r (3/2)
Add ½(b)2 to both sides
b
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
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r - (5/4) (7/4) 2/4 ½
and r - (5/4) - (7/4) -12/4 - 3
r ½ , - 3
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Ex Solve 3p 5 (p 1)(p 2)
Is this a quadratic equation? FOIL the RHS
3p 5 p2 2p p 2
Collect all terms
3p 5 p2 3p 2
p2 6p 7 0
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
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The Quadratic Formula

• Consider a quadratic equation of the form ax2
  bx c 0 for a nonzero
• Completing the square

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The Quadratic Formula

• Solutions to ax2 bx c 0 for a nonzero are

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Ex Use the Quadratic Formula to solve x2 7x
6 0
1
7
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Recall For quadratic equation ax2 bx c 0,
the solutions to a quadratic equation are given
by
Identify a, b, and c in ax2 bx c 0
a b c
1
7
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Now evaluate the quadratic formula at the
identified values of a, b, and c
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x ( - 7 5)/2 - 1 and x (-7 5)/2 - 6
x - 1, - 6
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Ex Use the Quadratic Formula to solve
2m2 m 10 0
1
2
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Recall For quadratic equation ax2 bx c 0,
the solutions to a quadratic equation are given
by
Identify a, b, and c in am2 bm c 0
a b c
2
1
- 10
Now evaluate the quadratic formula at the
identified values of a, b, and c
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m ( - 1 9)/4 2 and m (-1 9)/4 - 5/2
m 2, - 5/2
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Any questions . . .
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