Question

Find the value of n such that x^2-11x=n=0 is a perfect square trinomial.

Answer 1

You want to find \(n\) and \(k\) such that
\[x^2-11x+n=(x-k)^2=0\]
To find the values of \(n\) and \(k\) for this to be true, expand the middle expression and set it equal to the first:
\[x^2-2kx+k^2=x^2-11x+n\]
Matching up the coefficients of the same-power terms, you get
\[-2k=-11\quad\text{and}\quad k^2=n\]