# How do you find the c that makes the trinomial x^2+22x+c a perfect square?

May 15, 2017

$c = 121$

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

#### Explanation:

This is a process called 'Completing the Square' and does exactly what the name implies...

To complete means to add what is missing

You are trying to create a perfect square, in this case the square of a binomial.

In $1 {x}^{2} + \textcolor{red}{b} x + c , \text{ }$ if this is a perfect square there is always a specific relationship between $b \mathmr{and} c$....

'Half of $\textcolor{red}{b}$, squared, will give the value of $c$'

This is $c = {\left(\frac{\textcolor{red}{b}}{2}\right)}^{2}$

In 1x^2+ color(red)(22)x + ???" "rarr ??? = (color(red)(22)/2)^2 = 11^2 =121

The trinomial will therefore be ${x}^{2} + 22 x + 121$, which factorises as

${\left(x + 11\right)}^{2}$

Note that to do this, the coefficient of ${x}^{2}$ must be $1$