How do you find the value of c that makes $x^{2} - 19 x + c$ $x^2-19x+c$ into a perfect square?

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How do you find the value of c that makes $x^{2} - 19 x + c$ $x^2-19x+c$ into a perfect square?

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Answer 1 · 2017-07-07T13:58:06

$c = - {(\frac{19}{2})}^{2}$ $c=-(19/2)^2$

Explanation:

You square half of the 'b' term. This is because if you take the general binomial ${(a + b)}^{2}$ $(a+b)^2$ and expand it you get $a^{2} + 2 a b + b^{2}$ $a^2+2ab+b^2$ . In your quadratic, we can view 'x' as the 'a' and -19 as '2a'. Thus, if -19 = 2b we divide by 2 to give us $b = - \frac{19}{2}$ $b=-19/2$ and finally, we square it to give us the b squared.

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How do you find the value of c that makes $x^{2} - 19 x + c$ $x^2-19x+c$ into a perfect square?

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