# What number should be added to both sides of the equation to complete the square? x2 – 6x = 5

What number should be added to both sides of the equation to complete the square?

x2 – 6x = 5

## This Post Has 8 Comments

1. emanihackle9 says:

9

Step-by-step explanation:

Given

x² - 6x = 5

To complete the square

add ( half the coefficient of the x- term )² to both sides

x² + 2(- 3)x + 9 = 5 + 9, that is

(x - 3)² = 14

2. alondra6190 says:

9 should be added to complete the square

Step-by-step explanation:

1. Write the equation in a way that the constants are in the right side while the terms with x are on the left.

x^2 – 6x = 5

2. Make sure that the coefficient of the x^2 term is 1.

x^2 – 6x = 5

3. Adding a term to both sides that will complete the square in the left side. This is done by dividing the coefficient of the x term by 2 and squaring it. Note: The same amount should be added to the right side to balance the equation.

x^2 – 6x + 9 = 5 + 9

(x-3)^2 = 14

Therefore, 9 should be added to both sides to complete the square.

3. tifftiff22 says:

9

Step-by-step explanation:

4. johns2380 says:

I think 9 should be added to both sides.

cofficient of x = 6

half of it = 6/2 = 3

square the 3

to give 3 squared = 3*3 9

5. ashiteru123 says:

9

Step-by-step explanation:

To make it a Perfect Square Trinomial, you can square root 9 and multiply it to get 6, therefore 9 is correct

6. marissakirk8785 says:

7. Mfcarla says:

First of all, it is better to note that x2 is not equal to x² .In mathematics x2 is equal to 2nd x or 2x i.e. 2 times x=2*x=x*2.

I don't know what this question means ! But if it means to just confine x2 and -6x in a square or to make them part of a square then it could be done as follows.

Assume x2 as x² as per your needs as case may be.

Now use the identity

(a² - 2a×b + b²) =(a - b )²

Here , in LHS

a appears twice as in a² and in 2a×b.

So for given expression (x2 – 6x = 5) or (x² – 6x = 5) or (x² - 6x -5 =0)

Assume x=a

Then,

6x =2*x*3 =2*a*b

=> 3 = b

Thus

(x² - 6x -5 =0)

=> x² - 2*x*3 + 3² -3² -5 =0

=> ( x² - 2*x*3 + 3² ) - 3² -5 =0

=> (x - 3)² - 3² = 5

=> (x - 3)² = 5 + 3²

=> (x - 3)² = 5 + 9 = 14 = ( ±√(14) )²

Now comparing

(x² – 6x = 5) and (x - 3)² = 5 + 9,

it can be said that number 9 should be added to both sides of (x² – 6x = 5) or (x² – 6x) = ( 5 ) to make ( x² ) and (-6x ) as the constituting parts of the square (x-3)² .

I hope it helps.

Thank you  for patiently reading it.

8. coryowens44 says:

9

Step-by-step explanation:

x^2 - 6x = 5

Take the coefficient of x

-6

Divide it by 2

-6/2 = -3

Square it

(-3)^2 =9