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

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Answer 1 · 2017-04-13T17:16:02

$169/4$

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

The perfect square would be $(x-13/2)^2$

Answer 2 · 2017-04-13T17:17:04

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

This is the process called 'completing the square'

A quadratic trinomial is in the form $ax^2 +bx+c$

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

In this case $b = -13$

$x^2 -13x + ((-13)/2)^2$ is a perfect square

$x^2 -13x + 42.25$

$= (x-6.5)^2$

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