# For what value of k is x^2-21x+k a perfect square trinomial?

Mar 26, 2017

$k = \frac{441}{4} = 110.25$

#### Explanation:

Suppose:

${x}^{2} - 21 x + k = {\left(x + e\right)}^{2}$

$\textcolor{w h i t e}{{x}^{2} - 21 x + k} = {x}^{2} + 2 e x + {e}^{2}$

Equating coefficients, we have:

$2 e = - 21$

So:

$e = - \frac{21}{2}$

and

$k = {e}^{2} = {\left(- \frac{21}{2}\right)}^{2} = \frac{441}{4} = 110.25$