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Solving Quadratic
Equation Using
Completing the Square
M s . G . M a r t i n

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To solve the quadratic equation ax2 + bx + c = 0 by
completing the square, follow these steps.
1. Isolate the constant term to the right side of the equation.
2. Complete the square of the left side of the equation.
3. Express the left side of the equation as a square of the
binomial.
4. Apply the Square Root Property.
5. Solve for the roots.

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Example 1. Find the solution set of x2 – 18x = -17.
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Steps Solution
1. Isolate the terms with x,
or write in the form of ax2 + bx = c
x2 -18x = -17
2. Get the square of one-half the
coefficient of x.
x2 – 18x = -17
=
−18
2
= -9
= (-9)2
= 81
3. Complete the square of the left side of
the equation by adding 81 to both sides
of x2 – 18x = -17
x2 – 18x = -17
x2 – 18x + 81 = -17 + 81

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Steps Solution
3. Simplify the result in step 2.
Express the left side of the equation
as a square of the binomial.
x2 – 18x + 81 = -17 + 81
(x – 9)2 = 64
4. Apply the Square Root Property.
(square both sides)
(x – 9)2 = 64
x – 9 = ± 64
x – 9 = ±8
5. Solve for the roots x – 9 = ±8
x – 9 = 8 and x – 9 = -8
x = 8 + 9 and x = -8 + 9
x = 17 and x = 1

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Checking:
x = 1
x2 – 18x = -17
(1)2 – 18 (1) = - 17
1 – 18 = -17
-17 = -17
x = 17
x2 – 18x = -17
(17)2 – 18(17) = -17
289 – 306 = -17
-17 = -17

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Example 2. Solve x2 + 10x – 11 = 0.
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Steps Solution
1. Isolate the terms with x,
or write in the form of ax2 + bx = c
x2 + 10x = 11
2. Get the square of one-half the
coefficient of x.
x2 + 10x = 11
=
10
2
= (5)2
= 25
3. Complete the square of the left side
of the equation by adding 25 to both
sides of x2 + 10x = 11.
x2 + 10x = 11
x2 + 10x + 25 = 11 + 25

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Steps Solution
3. Simplify the result in step 2.
Express the left side of the equation
as a square of the binomial.
x2 + 10x + 25 = 11 + 25
(x + 5)2 = 36
4. Apply the Square Root Property.
(square both sides)
(x + 5)2 = 36
x + 5 = ± 36
x + 5 = ±6
5. Solve for the roots x + 5 = ±6
x + 5 = 6 and x + 5 = -6
x = 6 – 5 and x = -6 – 5
x = 1 and x = - 11

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Example 3. Solve x2 = - 24 + 10x.
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Solution:
x2 – 10x = -24
=
−𝟏𝟎
𝟐
= -5
= (5)2 = 25
x2 – 10x + 25 = -24 + 25
(x – 5) (x - 5) = 1
(x – 5)2 = 1
x – 5 = 𝟏
x – 5 = ±1
Finding the roots:
x – 5 = 1 and x – 5 = -1
x = 1 + 5 and x = -1 + 5
x = 6 and x = 4

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Thank You!