How do you find the value of c that makes $x^{2} - 13 x + c$ $x^2-13x+c$ into a perfect square?

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How do you find the value of c that makes $x^{2} - 13 x + c$ $x^2-13x+c$ into a perfect square?

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Answer 1 · 2017-04-13T17:16:02

$\frac{169}{4}$ $169/4$

Explanation:

Half the coefficient of x which would be $- \frac{13}{2}$ $-13/2$ and then square it which would be $\frac{169}{4}$ $169/4$ which 'c' should be equal to.

The perfect square would be ${(x - \frac{13}{2})}^{2}$ $(x-13/2)^2$

Answer 2 · 2017-04-13T17:17:04

$x^{2} - 13 x + 42.25$ $x^2 -13x + 42.25$ is a perfect square.

Explanation:

This is the process called 'completing the square'

A quadratic trinomial is in the form $a x^{2} + b x + c$ $ax^2 +bx+c$

To make a perfect square, add on ${(\frac{b}{2})}^{2}$ $(b/2)^2$ as the $c$ $c$ term

In this case $b = - 13$ $b = -13$

$x^{2} - 13 x + {(\frac{- 13}{2})}^{2}$ $x^2 -13x + ((-13)/2)^2$ is a perfect square

$x^{2} - 13 x + 42.25$ $x^2 -13x + 42.25$

$= {(x - 6.5)}^{2}$ $= (x-6.5)^2$

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How do you find the value of c that makes $x^{2} - 13 x + c$ $x^2-13x+c$ into a perfect square?

2 Answers

Explanation:

Explanation:

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How do you find the value of c that makes x^2-13x+cx2−13x+cx^2-13x+c into a perfect square?

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How do you find the value of c that makes $x^{2} - 13 x + c$ $x^2-13x+c$ into a perfect square?