Completing the Square – Method & Equation
The completing the square method means that we transform a quadratic equation in the usual form of x2 + 2bx + c and put it in this format: (x + b)2 – b2 + c
So, the completing the square equation is:
x2 + 2bx + c = (x + b)2 – b2 + c
Completing the Square Equation – Exercises
Instructions: Use the completing the square method to write the following quadratic equations in the completed square form. Then solve for x.
1) x2 + 6x + 4 = 0
2) x2 – 8x + 1 = 0
3) x2 – 4x + 15 = 0
4) 2x2 + 8x – 3 = 0
5) 3x2 – 6x – 7 = 0
Completing the Square Equation – Answers
1) The correct answer is: \(\text{x = }-\text{3} \pm \sqrt{5}\)
STEP 1: Expand the equation to identify variables b and c.
Our equation x2 + 6x + 4 = 0 is in the format x2 + 2bx + c.
x2 + (2 × 3 × x) + 4 = x2 + 2bx + c
b = 3 and c = 4
STEP 2: Put the equation in the format (x + b)2 – b2 + c
(x + 3)2 – 32 + 4
STEP 3: Perform the operations to simplify.
(x + 3)2 – 32 + 4 = 0
(x + 3)2 – 9 + 4 = 0
(x + 3)2 – 5 = 0
(x + 3)2 – 5 + 5 = 0 + 5
(x + 3)2 = 5
STEP 4: Take the square root of each side and simplify to solve.
(x + 3)2 = 5
\(\sqrt{(x + 3)^2} = \pm \sqrt{5}\)
\(\text{(x + 3) = } \pm \sqrt{5}\)
\(\text{x + 3} – 3 = \pm \sqrt{5} – \text{3}\)
\(\text{x = }-\text{3} \pm \sqrt{5} \)
2) The correct answer is: \(\text{x = }\text{4} \pm \sqrt{15} \)
STEP 1: Expand the equation to identify variables b and c.
Our equation x2 – 8x + 1 = 0 is in the format x2 + 2bx + c.
x2 + (2 × –4 × x) + 1 = x2 + 2bx + c
b = –4 and c = 1
STEP 2: Put the equation in the format (x + b)2 – b2 + c
(x – 4)2 – (–4)2 + 1
STEP 3: Perform the operations to simplify.
(x – 4)2 – (–4)2 + 1 = 0
(x – 4)2 – 16 + 1 = 0
(x – 4)2 – 15 = 0
(x – 4)2 – 15 + 15 = 0 + 15
(x – 4)2 = 15
STEP 4: Take the square root of each side and simplify to solve.
(x – 4)2 = 15
\(\sqrt{(x – 4)^2} = \pm \sqrt{15}\)
\(\text{(x} – 4) = \pm \sqrt{15}\)
\(\text{x} – 4 + 4 = \pm \sqrt{15} + \text{4}\)
\(\text{x = 4} \pm \sqrt{15} \)
3) The correct answer is: \(\text{x = 2} \pm \sqrt{-11} \)
STEP 1: Expand the equation to identify variables b and c.
Our equation x2 – 4x + 15 = 0 is in the format x2 + 2bx + c.
x2 + (2 × –2 × x) + 15 = x2 + 2bx + c
b = –2 and c = 15
STEP 2: Put the equation in the format (x + b)2 – b2 + c
(x – 2)2 – (–2)2 + 15
STEP 3: Perform the operations to simplify.
(x – 2)2 – (–2)2 + 15 = 0
(x – 2)2 – 4 + 15 = 0
(x – 2)2 + 11 = 0
(x – 2)2 + 11 – 11 = 0 – 11
(x – 2)2 = –11
STEP 4: Take the square root of each side and simplify to solve.
(x – 2)2 = –11
\(\sqrt{(x – 2)^2} = \pm \sqrt{-11}\)
\(\text{(x} – 2) = \pm \sqrt{-11}\)
\(\text{x} – 2 + 2 = \pm \sqrt{-11} + \text{2}\)
\(\text{x = 2} \pm \sqrt{-11} \)
4) The correct answer is: \(\text{x =} – 2 \pm \sqrt{\frac{11}{2}} \)
STEP 1: Expand the equation to identify variables a, b, and c.
\(\text{ax}^{2} + bx + c = a\bigg(x + \frac{b}{2a}\bigg)^{2}+ d\)
Our equation 2x2 + 8x – 3 = 0 is in the format ax2 + bx + c.
2x2 + (8 × x) – 3 = 0
a = 2 b = 8 c = –3
STEP 2: Expand the equation in squared form to calculate variable d.
(i) Divide variable b by variable a: 8 ÷ 2 = 4
(ii) Divide the previous calculation by 2: 4 ÷ 2 = 2
(iii) Use the result from (ii) as the second term in the parentheses to factor the quadratic: 2(x + 2)2
(iv) Simplify to calculate d:
2(x + 2)2 =
2(x + 2)(x + 2) =
[2(x × x) + (x × 2) + (2 × x) + (2 × 2)] =
[2(x2 + 4x + 4)] =
2x2 + 8x + 8 We deduct this number from c to get d.
c = –3 d = –3 – 8 = –11
STEP 3: Put the equation in the format \(\text{a}\bigg(x + \frac{b}{2a}\bigg)^{2}+ d\)
a = 2 b = 8 d = –11
\(\text{2}\bigg(x + \frac{8}{2 \times 2}\bigg)^{2}+ -11\)
STEP 4: Perform the operations to simplify.
\(\text{2}\bigg(x + \frac{8}{2 \times 2}\bigg)^{2}+ -11\)
\(\text{2}\bigg(x + \frac{8}{4}\bigg)^{2}+ -11\)
2(x + 2)2 – 11 = 0
2(x + 2)2 – 11 + 11 = 0 + 11
2(x + 2)2 = 11
[2(x + 2)2 ÷ 2] = 11 ÷ 2
\(\text{(x + 2)}^{2} = \frac{11}{2}\)
STEP 5: Take the square root of each side and simplify to solve.
\(\text{(x + 2)}^{2} = \frac{11}{2}\)
\(\sqrt{(x + 2)^2} = \pm \sqrt{\frac{11}{2}}\)
\(\text{(x} + 2) = \pm \sqrt{\frac{11}{2}}\)
\(\text{x} + 2 – 2 = \pm \sqrt{\frac{11}{2}} – \text{2}\)
\(\text{x =} – 2 \pm \sqrt{\frac{11}{2}} \)
5) The correct answer is: \(\text{x =} 1 \pm \sqrt{\frac{10}{3}} \)
STEP 1: Expand the equation to identify variables a, b, and c.
\(\text{ax}^{2} + bx + c = a\bigg(x + \frac{b}{2a}\bigg)^{2}+ d\)
Our equation 3x2 – 6x – 7 = 0 is in the format ax2 + bx + c.
3x2 + (–6 × x) – 7 = 0
a = 3 b = –6 c = –7
STEP 2: Expand the equation in squared form to calculate variable d.
(i) Divide variable b by variable a: –6 ÷ 3 = –2
(ii) Divide the previous calculation by 2: –2 ÷ 2 = –1
(iii) Use the result from (ii) as the second term in the parentheses to factor the quadratic: 3(x – 1)2
(iv) Simplify to calculate d:
3(x – 1)2 =
3(x – 1)(x – 1) =
[3(x × x) + (x × –1) + (–1 × x) + (–1 × –1)] =
[3(x2 – 2x + 1)] =
3x2 – 6x + 3 We deduct this number from c to get d.
c = –7 d = –7 – 3 = –10
STEP 3: Put the equation in the format \(\text{a}\bigg(x + \frac{b}{2a}\bigg)^{2}+ d\)
a = 3 b = –6 d = –10
\(\text{3}\bigg(x + \frac{-6}{2 \times 3}\bigg)^{2}+ -10\)
STEP 4: Perform the operations to simplify.
\(\text{3}\bigg(x + \frac{-6}{2 \times 3}\bigg)^{2}+ -10\)
\(\text{3}\bigg(x + \frac{-6}{6}\bigg)^{2}+ -10\)
3(x – 1)2 – 10 = 0
3(x – 1)2 – 10 + 10 = 0 + 10
3(x – 1)2 = 10
[3(x – 1)2 ÷ 3] = 10 ÷ 3
\(\text{(x – 1)}^{2} = \frac{10}{3}\)
STEP 5: Take the square root of each side and simplify to solve.
\(\text{(x – 1)}^{2} = \frac{10}{3}\)
\(\sqrt{(x – 1)^2} = \pm \sqrt{\frac{10}{3}}\)
\(\text{x} – 1 = \pm \sqrt{\frac{10}{3}}\)
\(\text{x} – 1 + 1 = \pm \sqrt{\frac{10}{3}} + \text{1}\)
\(\text{x =} 1 \pm \sqrt{\frac{10}{3}} \)
Completing the Square Method – Example
Let’s look at an example problem that illustrates the completing the square method.
Write x2 + 6x + 20 in the form (x + b)2 – b2 + c
STEP 1: Identify variables b and c.
Our equation x2 + 6x + 20 is in the format x2 + 2bx + c.
Expand the equation to identify b:
x2 + (2 × 3 × x) + 20 = x2 + 2bx + c
So, b = 3 and c = 20
STEP 2: Put the equation in the format (x + b)2 – b2 + c
When b = 3 and c = 20
(x + 3)2 – 32 + 20
STEP 3: Perform the operations to simplify.
(x + 3)2 – 32 + 20 =
(x + 3)2 – 9 + 20 =
(x + 3)2 + 11
STEP 4: You may check your answer by performing the multiplication using the FOIL method.
(x + 3)2 + 11 =
(x + 3)(x + 3) + 11 =
[(x × x) + (x × 3) + (3 × x) + (3 × 3)] + 11 =
x2 + 3x + 3x + 9 + 11 =
x2 + 6x + 20 (which was our original equation)
Completing the Square – Solving for x
Let’s look at our equation from step 3 above.
To solve for x, first we have to move the spare number to the other side.
(x + 3)2 + 11 = 0
(x + 3)2 + 11 – 11 = –11
(x + 3)2 = –11
Then take the square root of each side and simplify to solve.
(x + 3)2 = –11
\(\sqrt{(x + 3)^2} = \pm \sqrt{-11}\)
\(\text{(x + 3) = } \pm \sqrt{-11}\)
\(\text{x + 3} – 3 = \pm \sqrt{-11} – \text{3}\)
\(\text{x = } \pm \sqrt{-11} – \text{3}\)
\(\text{x = }- \text{3} \pm \sqrt{-11}\)
We always need to put the plus-minus symbol in front of the square root symbol on the right side of the equation when we solve for x.
The square root of a negative number is always an imaginary number, so we would be to use variable i to simplify the problem further.
Completing the Square – Advanced Problems
If you have an advanced math part on your algebra test, you will need to know the following formula for the advanced completing the square method:
\(\text{ax}^{2} + bx + c = a\bigg(x + \frac{b}{2a}\bigg)^{2}+ d\)
You have to do a separate calculation to determine what value to subtract from variable c in order to get variable d.
Completing the Square and Quadratic Graphs
Quadratic equations appear as curves when plotted on a graph.
All quadratic graphs will have a minimum point at the bottom of the curve.
Completing the square can also be used in order to find the x and y coordinates of the minimum value of a quadratic equation on a graph.
We cover how to graph quadratics in more depth in our graphing posts.
You may also want to try our other free algebra problems.
