which value of k makes p^2-8p+k a perfect square?

Answer 1

let, p^2-8p+k =0
then,we have to find two numbers whose sum is -8 and product is k and then by solving the equation, we will get 0. so,such numbers are -4 and-4
by expanding the equation, we have
p^2 (-4-4)p +(-4×-4)=0
p^2 -4p -4p +16=0
p(p-4) -4(p-4)=0
(p-4) (p-4)=0
p-4=0
p=4
by putting the value of p in equation, we have
=(4^2) -(8×4) +(16)
=(16) -(32) +(16)
=32-32
=0.
so,the value of k is 16.