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#