MAT1033-5
Solving Quadratic Equations
METHOD 1: FACTORING
Key Facts:
Steps:
1.
2.
3.
4.
Set equation equal to zero.
Factor.
Set the factors equal to zero.
Solve these newly formed equations.
Example 1: Solve x2 – 12 = x
1. Set equation equal to zero.
x 2 -x-12=0
2. Factor.
(x-4)(x+3)=0
3. Set the factors equal to zero.
x-4=0;x+3=0
4. Solve these newly formed equations.
x=4;x=-3
Example 2: Solve y = x2 – 10x + 25
1. Set equation equal to zero.
x 2 -10x+25=0
2. Factor.
(x-5)(x-5)=0
3. Set the factors equal to zero.
x-5=0; x-5=0
4. Solve these newly formed equations.
x=5;x=5
1. Quadratic is another name for a
polynomial of the 2nd degree; i.e.,
‘2’ is the highest exponent. A
quadratic equation has the form ax²
+ bx + c = 0
2. The root of a quadratic equation is
its solution, and a quadratic always
has two roots, real or complex. A
double root occurs when the two
roots are equal.
3. The graph of a quadratic is always a
parabola, and is similar to the
following graph of y = 3x 2 − 4 :
30
20
10
0
‐3 ‐2 ‐1 0 1 2 3
‐10
4. The following identities are helpful in
solving quadratic equations:
a. If m x n = 0, then either m = 0
or n = 0 (or both)
b. If A2=C, then A= √
MAT1033-5
The "two" roots are 5 and 5. In this case, ‘5’ is called a double
root. At a double root, the graph does not cross the x-axis. It
just touches it. A double root occurs when the quadratic is a
perfect square trinomial: x² ±2ax + a²; that is, when it is the
square of a binomial: (x ± a)².
METHOD 2: COMPLETING THE SQUARE
Example 3: Solve 16x2 – 8x – 63 = 0 (a=16; b= -8; c= -63)
After a short amount of time the student should find it is not possible to factor this
polynomial the way it was done in Example 1. In cases like this another method will be
used: Completing the Square. This means to make the quadratic into a perfect square
trinomial, i.e. the form a² + 2ab + b² = (a + b) ². To accomplish this, perform the
following:
Steps:
1. Reduce the coefficient of the squared term to +1 by dividing both sides by ‘a’.
(Reason: This technique is valid only if the coefficient of x2 is +1)
16x2 – 8x – 63 = 0
(Divide by 16)
x2 – 8/16 x – 63/16 = 0
x2 – ½ x – 63/16 = 0
2. Move the c term (constant) to the right side of the equation
x2 – ½ x + ___= 63/16 + ___
3. Take ½ of b (coefficient of single powered term) and then square it.
(½ * b)2 =( ½* ½)2 = ¼2=1/16
4. Add this number to both sides of the equation.
x2 – ½ x + 1/16 = 63/16 + 1/16
x2 – ½ x + 1/16 = 64/16
(x- ¼)2 = 4
5. Now solve by extracting roots.
x - ¼ = ±2
x = ¼ ±2
x = 9/4; =7/4
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Example 4: Solve x2 + 6x + 2 = 0 (a=1; b= 6; c= 2)
Steps:
1. Reduce the coefficient of the squared term to +1 by dividing both sides of the
equation by a.
Not necessary; coefficient of squared term is +1
2. Move the c term (constant) to the right side of the equation
x2 + 6x + __ = -2 + ___
3. Take ½ of b (coefficient of single powered term) and then square it.
(½ * b)2 =( ½* 6)2 = 32 = 9
4. Add this number to both sides of the equation.
x2 + 6x + 9 = -2 + 9
x2 + 6x + 9 = + 7
(x + 3)2 = 7
5. Now solve by extracting roots.
x + 3 = ±√7
x = -3 ±√7
The solution is the conjugate pair, −3 + 7 , −3 − 7
MAT1033-5
METHOD 3: USING THE QUADRATIC FORMULA
If ax² + bx + c = 0, then
To prove this, we will complete the square. But to do that, the coefficient of x² must be +1
Therefore, we will divide both sides of the original equation by a:
ax 2 + bx + c = 0
x2 +
bx c
+ =0
a a
x2 +
bx
c
=−
a
a
2
⎛1 b ⎞
b2
⎜⎜⎜ * ⎟⎟⎟ = 2
⎜⎝2 a ⎟⎠
4a
x2 +
bx
b2
c
b2
+ 2 =− + 2
a 4a
a 4a
2
⎛
⎞
⎜⎜x + b ⎟⎟ = −4ac + b
⎜⎜⎝
2a ⎠⎟⎟
4a 2
⎛
⎞
⎜⎜x + b ⎟⎟ −4ac + b
⎜⎜⎝
2a ⎠⎟⎟
4a 2
x=
−4ac + b b
−
2a
4a 2
x=
−b
−4ac + b
±
2a
2a
x=
−b ± b2 − 4ac
2a
This is the quadratic formula
MAT1033-5
Example 5: Solve 3x² + 5x − 8 = 0 using the quadratic formula.
Solution. We have: a = 3, b = 5, c = −8.
Therefore, according to the formula:
x =
x =
−b ±
−5 ±
b 2 − 4a c
2a
52 − 4 * 3 * − 8
2*3
x =
− 5 ± 121
6
x =
− 5 ± 11
6
x= + 1 ; -
8
3
Example 6: Solve x² - 5x + 5 = 0 using the quadratic formula.
a = 1, b = −5, c = 5
−(− 5) ± − 5 2 − 4 * 1 * 5
x =
2*1
x =
5± 5
2
x =
5± 5
2
MAT1033-5
EXERCISES
1. Find the roots of each quadratic by factoring.
a) x² − 3x + 2
b) x² + 7x + 12
c) x² + 3x − 10
d) x² − x − 30
e) 2x² + 7x + 3
f) 3x² + x − 2
g) x² + 12x + 36
h) x² − 2x + 1
2. Find the roots of each quadratic.
a) x² − 5x
b) x² + x
c) 3x² + 4x
d) 2x² − x
3. Find the roots of each quadratic.
a) x² − 3
b) x² − 25
c) x² − 10
4. Solve each equation for x.
a) x² = 5x − 6
b) x² + 12 = 8x
c) 3x² + x = 10
d) 2x² = x
5. Solve each quadratic equation by completing the square.
a) x² − 2x − 2 = 0
b) x² + 4x − 6 = 0
c) x² − 4x + 13 = 0
d) x² + 6x + 29 = 0
e) x² − 5x – 5 = 0
f) x² + 3x + 1
= 0
6. Solve the following using the quadratic formula:
a) -3(x+1) = (2x – 1)2
b) y2 + 3y + 2 = 0
c) 3t2 = 7t + 6
d) z2 + 4z +1 = 0
e) 4p2 + 16p = -11
MAT1033-5
ANSWERS
1. Find the roots of each quadratic by factoring.
a) x² − 3x + 2
b) x² + 7x + 12
(x − 1)(x − 2)
(x + 3)(x + 4)
x = 1 or 2.
x = −3 or −4.
Notice that we use the conjunction "or," because
x takes on only one value at a time.
c) x² + 3x − 10
d) x² − x − 30
(x + 5)(x − 2)
(x + 5)(x − 6)
x = −5 or 2.
x = −5 or 6.
e) 2x² + 7x + 3
f) 3x² + x − 2
(2x + 1)(x + 3)
(3x − 2)(x + 1)
x = − ½ or −3.
x = 2/3
g) x² + 12x + 36
or −1.
h) x² − 2x + 1
(x + 6)²
(x − 1)²
x = −6, −6.
x = 1, 1.
A double root.
A double root.
2. Find the roots of each quadratic.
a) x² − 5x
b) x² + x
x(x − 5)
x(x + 1)
x = 0 or 5.
x = 0 or −1.
MAT1033-5
c)
3x² + 4x
d) 2x² − x
x(3x + 4)
x(2x − 1)
x = 0 or −4/3
x = 0 or ½
3. Find the roots of each quadratic.
a) x² − 3
b) x² − 25
x² = 3
x=±
3.
c) x² − 10
(x + 5)(x − 5)
(x +
10 )(x
x = ±5.
x=±
−
10 )
10 .
4. Solve each equation for x.
a) x² = 5x − 6
b)
x² + 12 = 8x
x² − 5x + 6 = 0
x² − 8x + 12 = 0
(x − 2)(x − 3) = 0
(x − 2)(x − 6) = 0
x = 2 or 3.
x = 2 or 6.
d)
c) 3x² + x = 10
2x² = x
3x² + x − 10 = 0
2x² − x = 0
(3x − 5)(x + 2) = 0
x(2x − 1) = 0
x = 5/3 or − 2.
x = 0 or 1/2
5. Solve each quadratic equation by completing the square.
a) x² − 2x − 2
= 0
x² − 2x
= 2
x² + 4x
x² − 2x + 1
= 2+1
x² + 4x + 4 = 6 + 4
(x − 1)²
= 3
(x + 2)²
= 10
x+2
= ±
x
= −2 ±
x−1
=
±
x
=
1±
b) x² + 4x − 6 = 0
3
3
= 6
3
3
MAT1033-5
c) x² − 4x + 13 = 0
x² − 4x
d) x² + 6x + 29 = 0
−13
x² + 6x
x² − 4x + 4 =
−13 + 4
x² + 6x + 9 = −29+ 9
(x − 2)²
=
−9
(x + 3)²
= −20
x−2
=
±3i
x+3
=±
x
=
2 ± 3i
e) x² − 5x − 5
=
0
f) x² + 3x + 1 = 0
=
5
x² + 3x = −1
x² − 5x
=
x
=
−29
−20
= −3 ± 2i
5
x² − 5x + 25/4 = 5 + 25/4
x² + 3x+ 9/4 = −1+ 9/4
(x − 5/2)²
= 5 + 25/4
(x + 3/2)² =− 1 + 9/4
x − 5/2
=±
x
=
45
4
5 3
±
5
2 2
x + 3/2 = ±
x=
5
4
3 1
− ±
5
2 2
MAT1033-5
6. Solve the following using the quadratic formula:
a) -3(x+1) = (2x – 1)2
P ut in standard form ax 2 + bx + c = 0
− 3(x + 1) = (2 x − 1)2
− 3 x − 3 = (2 x − 1)(2 x − 1)
− 3x − 3 = 4 x 2 − 4 x + 1
0 = 4 x 2 − 4 x + 1 + 3x + 3
0 = 4x 2 − x + 4
a = 4; b = − 1; c = 4
Substitute into quadratic form ula
x =
x =
−(− 1) ±
1±
− 12 − (4 * 4 * 4)
8
− 63
8
1 ± 3i 7
x =
8
b) y2 + 3y + 2 = 0
y= {-1, -2}
c) 3t2 = 7t + 6
t= {-2/3, 3}
d) z2 + 4z +1 = 0
z= {-2+√3, -2 - √3}
e) 4p2 + 16p = -11
p={
√
,
√
}
−b ±
b 2 − 4a c
2a