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

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

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Answer 1 · 2017-05-15T07:26:47

$c = 121$ $c = 121$

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

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

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

This is $c = {(\frac{b}{2})}^{2}$ $c= (color(red)(b)/2)^2$

In $1 x^{2} + 22 x + ? ? ? \to ? ? ? = {(\frac{22}{2})}^{2} = 11^{2} = 121$ $1x^2+ color(red)(22)x + ???" "rarr ??? = (color(red)(22)/2)^2 = 11^2 =121$

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

${(x + 11)}^{2}$ $(x+11)^2$

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

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

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