How do you find the value of c that makes $x^{2} - 24 x + c$ $x^2-24x+c$ into a perfect square?

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How do you find the value of c that makes $x^{2} - 24 x + c$ $x^2-24x+c$ into a perfect square?

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Answer 1 · 2018-01-05T01:36:49

Divide the coefficient of $x$ $x$ (not $x^{2}$ $x^2$ ) by 2, and then square that.
$c = 144$ $c=144$ .

Explanation:

Divide the coefficient of $x$ $x$ (not $x^{2}$ $x^2$ ) by 2, and then square that.

So for $x^{2} - 24 x$ $x^2-24x$ , dividing $- 24$ $-24$ by $2$ $2$ gets us $- 12$ $-12$ .
Squaring $- 12$ $-12$ gets us $144$ $144$ ,

So $c = 144$ $c=144$ .

In general:

Consider the factoring identity

$\qquad(x-a)^2 = x^2 - 2ax+ a^2$

If we only have the middle term, $- 2ax$ , then dividing the middle term by $2x$ gets us

$\qquad\frac{-2ax}{2x} = -a$

(If we only take the coefficient on $x$ , we do not have to divide by $x$ .)

Then if we square that, we get

$\qquad(-a)^2 = a^2$

So as you can see, taking the coefficient on $x$ , dividing it by $2$ , and then squaring it will get us the magic constant that lets us complete the square.

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How do you find the value of c that makes $x^{2} - 24 x + c$ $x^2-24x+c$ into a perfect square?

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How do you find the value of c that makes $x^{2} - 24 x + c$ $x^2-24x+c$ into a perfect square?