Answer:
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.
