# How do you find the value of c that makes x^2-22x+c into a perfect square?

Nov 23, 2016

$c = 121$

#### Explanation:

We have

${x}^{2} - 22 x + c$

We want to make this into a perfect square, and suppose we choose the value of k such that:

${x}^{2} - 22 x + c = {\left(x + k\right)}^{2}$ (NB we expect k to -ve)
$\therefore {x}^{2} - 22 x + c = {x}^{2} + 2 k + {k}^{2}$

Comparing coefficient of $x$ we have

$2 k = - 22 \implies k = - 11$

And comparing constant coefficients we have:
$c = {k}^{2} \implies c = 121$

Hence If we choose $c = 121$ then we can write

${x}^{2} - 22 x + 121 = {\left(x - 11\right)}^{2}$