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

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Answer 1 · 2017-05-15T07:26:47

$c = 121$

Which gives $x^2 +22x+121 =(x+11)^2$

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 $1x^2 + color(red)(b)x + c," "$ if this is a perfect square there is always a specific relationship between $b and c$ ....

'Half of $color(red)(b)$ , squared, will give the value of $c$ '

This is $c= (color(red)(b)/2)^2$

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

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

$(x+11)^2$

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

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