Introduction
We can solve quadratic equations by transforming the
left side of the equation into a perfect square trinomial
and using square roots to solve. Previously, you may
have explored perfect square trinomials in the form
(ax)2 ± 2abx + b2 . In this lesson, we will use a modified
2
æ bö
2
version of this form, x + bx +
çè 2 ÷ø .
1
3A.2.6: Completing the Square
Key Concepts
• When the binomial (x + a) is squared, the resulting
perfect square trinomial is x2 + 2ax + a2.
• When the binomial (ax + b) is squared, the resulting
perfect square trinomial is a2x2 + 2abx + b2.
2
3A.2.6: Completing the Square
Key Concepts, continued
Completing the Square to Solve Quadratics
1. Make sure the equation is in standard form,
ax2 + bx + c.
2. Subtract c from both sides.
3. Divide each term by a to get a leading
coefficient of 1.
4. Add the square of half of the coefficient of the
x-term to both sides to complete the square.
5. Express the perfect square trinomial as the
square of a binomial.
6. Solve by using square roots.
3
3A.2.6: Completing the Square
Common Errors/Misconceptions
• neglecting to add the c term of the perfect square
trinomial to both sides
• not isolating x after squaring both sides
• forgetting that, when taking the square root, both the
positive and negative roots must be considered (±)
4
3A.2.6: Completing the Square
Guided Practice
Example 2
Solve x2 + 6x + 4 = 0 by completing the square.
5
3A.2.6: Completing the Square
Guided Practice: Example 2, continued
1. Determine if x2 + 6x + 4 is a perfect
square trinomial.
Take half of the value of b and then square the
result. If this is equal to the value of c, then the
expression is a perfect square trinomial.
2
2
æ bö
æ 6ö
ç 2÷ = ç 2÷ = 9
è ø
è ø
x2 + 6x + 4 is not a perfect square trinomial because
the square of half of 6 is not 4.
3A.2.6: Completing the Square
6
Guided Practice: Example 2, continued
2. Complete the square.
x2 + 6x + 4 = 0
x2
+ 6x = –4
x2 + 6x + 32 = –4 + 32
x2 + 6x + 9 = 5
Original equation
Subtract 4 from both
sides.
Add the square of half of
the coefficient of the
x-term to both sides to
complete the square.
Simplify.
7
3A.2.6: Completing the Square
Guided Practice: Example 2, continued
3. Express the perfect square trinomial as
the square of a binomial.
Half of b is 3, so the left side of the equation can be
written as (x + 3)2.
(x + 3)2 = 5
8
3A.2.6: Completing the Square
Guided Practice: Example 2, continued
4. Isolate x.
(x + 3)2 = 5
Equation
x+3= ± 5
Take the square root of
both sides.
x = -3 ± 5
Subtract 3 from both
sides.
9
3A.2.6: Completing the Square
Guided Practice: Example 2, continued
5. Determine the solution(s).
The equation x2 + 6x + 4 = 0 has two solutions,
x = -3 ± 5.
✔
10
3A.2.6: Completing the Square
Guided Practice: Example 2, continued
11
3A.2.6: Completing the Square
Guided Practice
Example 3
Solve 5x2 – 50x – 120 = 0 by completing the square.
12
3A.2.6: Completing the Square
Guided Practice: Example 3, continued
1. Determine if 5x2 – 50x – 120 = 0 is a
perfect square trinomial.
The leading coefficient is not 1.
First divide both sides of the equation by 5 so that
a = 1.
5x2 – 50x – 120 = 0
Original equation
x2 – 10x – 24 = 0
Divide both sides by 5.
13
3A.2.6: Completing the Square
Guided Practice: Example 3, continued
Now that the leading coefficient is 1, take half of the
value of b and then square the result. If the
expression is equal to the value of c, then it is a
perfect square trinomial.
2
2
æ bö
æ -10 ö
ç 2 ÷ = ç 2 ÷ = 25
è ø
è
ø
5x2 – 50x – 120 = 0 is not a perfect square trinomial
because the square of half of –10 is not –24.
14
3A.2.6: Completing the Square
Guided Practice: Example 3, continued
2. Complete the square.
x2 – 10x – 24 = 0
Equation
x2 – 10x = 24
Add 24 to both
sides.
x2 – 10x + (–5)2 = 24 + (–5)2
Add the square of
half of the coefficient
of the x-term to both
sides to complete
the square.
x2 – 10x + 25 = 49
Simplify.
3A.2.6: Completing the Square
15
Guided Practice: Example 3, continued
3. Express the perfect square trinomial as
the square of a binomial.
Half of b is –5, so the left side of the equation can be
written as (x – 5)2.
(x – 5)2 = 49
16
3A.2.6: Completing the Square
Guided Practice: Example 3, continued
4. Isolate x.
(x – 5)2 = 49
Equation
x - 5 = ± 49 = ± 7
Take the square root of
both sides.
x = 5±7
Add 5 to both sides.
x = 5 + 7 = 12 or
x = 5 – 7 = –2
Split the answer into two
separate equations and
solve for x.
17
3A.2.6: Completing the Square
Guided Practice: Example 3, continued
5. Determine the solution(s).
The equation 5x2 – 50x – 120 = 0 has two solutions,
x = –2 or x = 12.
✔
18
3A.2.6: Completing the Square
Guided Practice: Example 3, continued
19
3A.2.6: Completing the Square