Question 431533


x^2+6x-2 = (x+p)^2+q

=> x^2+6x-2= x^2 +p^2 + 2px + q

now, compare co-efficient of both sides....

co-efficient of x on LHS = 6, on RHS = 2p 

so, 6 = 2p

=> p = 3

similarly constant value on LHS is -2, on RHS is p^2 +q 

so, -2 = p^2 +q

=> -2 = 3^2 + q  (after putting the value of p)

=> -2 -9 = q

here q = -11


p = 3 and q = -11