Worked examples
The question
Question: solve x² + 6x = 7 by completing the square.
x2 + 6x = 7
1 Complete the square
We want the left side to be one bracket, squared.
A picture shows what is missing.
So we add 9 to fill the corner.
We do it on both sides, so they stay equal.
x2 + 6x + 9 = 7 + 9
Careful!
Many people add 9 only on the left.
Then the two sides are not equal any more.
Careful: add 9 on BOTH sides
2 Write it as a square
The full square is x + 3 on each side.
So x² + 6x + 9 is (x + 3)².
x2 + 6x + 9 = (x + 3)2
7 plus 9 is 16.
7 + 9 = 16
Now we have (x + 3)² = 16.
(x + 3)2 = 16
3 Square root of both sides
Something squared is 16.
So that something is the square root of 16, or minus it.
Both work, because minus times minus is plus.
( )² = k → ( ) = √k or −√k
The square root of 16 is 4, because 4 times 4 is 16.
x + 3 = 4 or x + 3 = −4
4 Get x alone
To remove the + 3, we take away 3 on both sides.
We do it in both equations.
x = 4 − 3 or x = −4 − 3
4 minus 3 is 1.
Minus 4 minus 3 is −7.
4 − 3 = 1, −4 − 3 = −7
So x = 1 or x = −7.
x = 1 or x = −7
5 Check
Let's check: put x = 1 into the left side.
x = 1: 12 + 6(1) = 1 + 6 = 7
We get 7, the same as the right side.
It works!
Now put x = −7 into the left side.
x = −7: (−7)2 + 6(−7) = 49 − 42 = 7
We get 7, the same as the right side.
So both answers are right!
Let's look back.
First: complete the square.
Then: write it as a square.
Then: square root of both sides.
Then: get x alone.
Last: check.
▶ Watch it solved
The question
Question: solve x² − 4x − 5 = 0 by completing the square.
x2 − 4x − 5 = 0
1 Move the −5
We want only the x parts on the left.
The − 5 is in the way.
So we add 5 on both sides.
x2 − 4x − 5 + 5 = 0 + 5
− 5 and + 5 make zero.
Cross them out.
0 plus 5 is 5.
0 + 5 = 5
Now we have x² − 4x = 5.
x2 − 4x = 5
2 Complete the square
We want the left side to be one bracket, squared.
A picture shows what is missing.
So we add 4 to fill the corner.
We do it on both sides, so they stay equal.
x2 − 4x + 4 = 5 + 4
Careful!
Many people add 4 only on the left.
Then the two sides are not equal any more.
Careful: add 4 on BOTH sides
3 Write it as a square
The full square is x − 2 on each side.
So x² − 4x + 4 is (x − 2)².
x2 − 4x + 4 = (x − 2)2
5 plus 4 is 9.
5 + 4 = 9
Now we have (x − 2)² = 9.
(x − 2)2 = 9
4 Square root of both sides
Something squared is 9.
So that something is the square root of 9, or minus it.
Both work, because minus times minus is plus.
( )² = k → ( ) = √k or −√k
The square root of 9 is 3, because 3 times 3 is 9.
x − 2 = 3 or x − 2 = −3
5 Get x alone
To remove the − 2, we add 2 on both sides.
We do it in both equations.
x = 3 + 2 or x = −3 + 2
3 plus 2 is 5.
Minus 3 plus 2 is −1.
3 + 2 = 5, −3 + 2 = −1
So x = 5 or x = −1.
x = 5 or x = −1
6 Check
Let's check: put x = 5 into the left side.
x = 5: 52 − 4(5) − 5 = 25 − 20 − 5 = 0
We get 0, the same as the right side.
It works!
Now put x = −1 into the left side.
x = −1: (−1)2 − 4(−1) − 5 = 1 + 4 − 5 = 0
We get 0, the same as the right side.
So both answers are right!
Let's look back.
First: move the −5.
Then: complete the square.
Then: write it as a square.
Then: square root of both sides.
Then: get x alone.
Last: check.
▶ Watch it solved
The question
Question: solve 2x² − 4x − 6 = 0 by completing the square.
2x2 − 4x − 6 = 0
1 Divide by 2
We want just x², not 2x².
So we divide every part by 2.
We do it on both sides, so they stay equal.
2, −4, −6: all divide by 2
Now we have x² − 2x − 3 = 0.
x2 − 2x − 3 = 0
2 Move the −3
We want only the x parts on the left.
The − 3 is in the way.
So we add 3 on both sides.
x2 − 2x − 3 + 3 = 0 + 3
− 3 and + 3 make zero.
Cross them out.
0 plus 3 is 3.
0 + 3 = 3
Now we have x² − 2x = 3.
x2 − 2x = 3
3 Complete the square
We want the left side to be one bracket, squared.
A picture shows what is missing.
So we add 1 to fill the corner.
We do it on both sides, so they stay equal.
x2 − 2x + 1 = 3 + 1
Careful!
Many people add 1 only on the left.
Then the two sides are not equal any more.
Careful: add 1 on BOTH sides
4 Write it as a square
The full square is x − 1 on each side.
So x² − 2x + 1 is (x − 1)².
x2 − 2x + 1 = (x − 1)2
3 plus 1 is 4.
3 + 1 = 4
Now we have (x − 1)² = 4.
(x − 1)2 = 4
5 Square root of both sides
Something squared is 4.
So that something is the square root of 4, or minus it.
Both work, because minus times minus is plus.
( )² = k → ( ) = √k or −√k
The square root of 4 is 2, because 2 times 2 is 4.
x − 1 = 2 or x − 1 = −2
6 Get x alone
To remove the − 1, we add 1 on both sides.
We do it in both equations.
x = 2 + 1 or x = −2 + 1
2 plus 1 is 3.
Minus 2 plus 1 is −1.
2 + 1 = 3, −2 + 1 = −1
So x = 3 or x = −1.
x = 3 or x = −1
7 Check
Let's check: put x = 3 into the left side.
x = 3: 2(3)2 − 4(3) − 6 = 18 − 12 − 6 = 0
We get 0, the same as the right side.
It works!
Now put x = −1 into the left side.
x = −1: 2(−1)2 − 4(−1) − 6 = 2 + 4 − 6 = 0
We get 0, the same as the right side.
So both answers are right!
Let's look back.
First: divide by 2.
Then: move the −3.
Then: complete the square.
Then: write it as a square.
Then: square root of both sides.
Then: get x alone.
Last: check.
▶ Watch it solved
