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

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Answer 1 · 2016-11-23T16:13:30

$c = 121$ $c=121$

We have

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

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

$x^{2} - 22 x + c = {(x + k)}^{2}$ $x^2 -22x + c = (x+k)^2$ (NB we expect k to -ve)
$:. x^2 -22x + c = x^2+2k+k^2$

Comparing coefficient of $x$ we have

$2k=-22 => k=-11$

And comparing constant coefficients we have:
$c=k^2 => c=121$

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

$x^2 -22x + 121 = (x-11)^2$

How do you find the value of c that makes x^2-22x+cx2−22x+cx^2-22x+c into a perfect square?