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### Solving Equations by Completing the Square

- You can solve quadratic equations by taking the square root of both sides of the equation. To do this, the quadratic expression must be a perfect square.
*Complete the square*to convert a quadratic expression into a perfect square. Take one half of the coefficient of the linear term, square it, and add this value to*both sides*of the equation.- If the coefficient of the quadratic term is not 1, you must first factor this coefficient out of the quadratic and linear term and then complete the square for the new quadratic expression which now has a coefficient of 1.

### Solving Equations by Completing the Square

Solve:

x

x

^{2}+ 4x + 18 = 2- x
^{2}+ 4x = − 16 - x
^{2}+ 4x + ( [b/2] )^{2}= − 16 + ( [b/2] )^{2} - x
^{2}+ 4x + [16/4] = − 16 + [16/4] - x
^{2}+ 4x + 4 = − 16 + 4 - (x + 2)
^{2}= − 12

x + 2 = ±√{12}

No Solution

No Solution

Solve:

y

y

^{2}+ 10y + 14 = 70- y
^{2}+ 10y = 56 - y
^{2}+ 10y + ( [b/2] )^{2}= 56 - y
^{2}+ 10y + [100/4] = 56 + 25 - ( y + 5 )
^{2}= 81 - y + 5 = √{81}
- y + 5 = ±9
- y + 5 = 9y = 4
- y + 5 = − 9y = − 13

y = { − 13,4}

Solve:

k

k

^{2}− 14k − 27 = 68- k
^{2}− 14k = 95 - k
^{2}− 14k + ( [b/2] )^{2}= 95 + ( [b/2] )^{2} - k
^{2}− 14k + [196/4] = 95 + [196/4] - k
^{2}− 14k + 49 = 95 + 49 - ( k − 7 )
^{2}= 144 - k − 7 = ±12
- k − 7 = 12k = 19

k − 7 = − 12k = − 5

Find the value of c to complete the square:

r

r

^{2}− 8r + c- r
^{2}− 8r + ( [b/2] )^{2} - r
^{2}− 8r + ( [( − 8)/2] )^{2} - r
^{2}− 8r + 16 - Factor:

( r + [(( − 8 ))/2] )^{2}

( r − 4 )

^{2}Find the value of c to complete the square:

g

g

^{2}− 5g + c- g
^{2}− 5g + ( [b/2] )^{2} - g
^{2}− 5g + [ [(( − 5 ))/2] ]^{2} - g
^{2}− 5g + [25/4] - Factor:

( g + [(( − 5 ))/2] )^{2}

( g − [25/4] )

^{2}Find the value of c to complete the square:

m

m

^{2}+ 10m + c- m
^{2}+ 10m + ( [b/2] )^{2} - m
^{2}+ 10m + [(10^{2})/4] - m
^{2}+ 10m + 25 - Factor:

( m + [6/2] )^{2} - ( m + [10/2] )
^{2}

( m + 5 )

^{2}Solve by completing the square:

n

n

^{2}− 6n − 9 = 11- n
^{2}− 6n = 20 - n
^{2}− 6n + ( [b/2] )^{2}= 20 + ( [b/2] )^{2} - n
^{2}− 6n + [6/4]^{2}= 20 + [6/4]^{2} - n
^{2}− 6n + 9 = 20 + 9 - n
^{2}− 6n + 9 = 29 - ( n + [b/2] )
^{2}= 29 - ( n + [(( − 6 ))/2] )
^{2}= 29 - ( n − 3 )
^{2}= 29 - n − 3 = ±√{29}

n = 3 ±√{29}

Solve by completing the square:

3x

3x

^{2}+ 9x + 4 = 16- 3x
^{2}+ 9x = 12 - x
^{2}+ 3x = 4 - x
^{2}+ 3x + ( [b/2] )^{2}= 4 + ( [b/2] )^{2} - x
^{2}+ 3x + [9/4] = 4 + [9/4] - x
^{2}+ 3x + [9/4] = 6[1/4] - x
^{2}+ [3/2] = ±√{6[1/4]}

x = − [3/2] ±√{6[1/4]}

Solve by completing the square:

v

v

^{2}− 12v + 3 = 39- v
^{2}− 12v = 36 - v
^{2}− 12v + [(( 12 )^{2})/4] = 36 + [(( 12 )^{2})/4] - v
^{2}− 12v + 36 = 72 - ( v + [( − 12)/2] )
^{2}= 72 - ( v − 6 )
^{2}= 72

v − 6 = ±√{72} v = 6 ±√{72}

Solve by completing the square:

7w

7w

^{2}− 56w + 16 = 37- 7w
^{2}− 56w = 21 - w
^{2}− 8w = 3 - w
^{2}− 8w + [(8^{2})/4] = 3 + [(8^{2})/4] - w
^{2}− 8w + 16 = 19 - ( w − 4 )
^{2}= 19 - w − 4 = ±√{19}

w = 4 ±√{19}

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Solving Equations by Completing the Square

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro 0:00
- Perfect Square Trinomials 0:15
- Example
- Completing the Square 4:55
- Example
- Completing the Square to Solve Equations 9:19
- Example
- When the Leading Coefficient is Not 1 13:17
- Example
- Example 1: Solve the Equation 15:05
- Example 2: Complete the Square 20:16
- Example 3: Solve by Completing the Square 22:31
- Example 4: Solve by Completing the Square 25:02

2 answers

Last reply by: John Joaneh

Mon Feb 1, 2016 3:39 PM

Post by John Joaneh on January 29, 2016

professor, im looking for videos on completing the square for quadratic equations. but i cant find any. maybe im not looking in the right place. do you know if anyone teaches that here?

1 answer

Last reply by: Dr Carleen Eaton

Sun Jan 31, 2016 3:09 PM

Post by John Joaneh on January 29, 2016

at the end of example 1 you said there is no solution. but cant you just factor out a negative 1 and get the square root of 16 (plus or minus 4), and then just solve like normal?

2 answers

Last reply by: Dr Carleen Eaton

Sat Jun 22, 2013 11:41 AM

Post by Taylor Wright on June 22, 2013

Can't you just divide b by 2 and then square it?

0 answers

Post by Dr Carleen Eaton on January 7, 2013

Sure. An example would be x^2 + 6x + 9 = 9. You would get (x+3)^2=9, or x+3 = +/- 3, which gives x=0 or x=-6. (Note that it would be easier to solve this by starting from the original equation and subtracting 9 from both sides to get x^2+6x=0 and then factoring.)

0 answers

Post by Anand Gundakaram on January 6, 2013

can x equal 0 ?