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

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.

homozygous. one is homozygous, while two is heterozygous.