What is the value of #c# that makes #x^2-15x+c# a perfect square trinomial?

2 Answers
Feb 24, 2017

Answer:

#c = (15/2)^2 = 225/4#

Explanation:

We find:

#(x+b/2)^2 = x^2+2(x)(b/2)+(b/2)^2 = x^2+bx+b^2/4#

So in order for #x^2+bx+c# to be a perfect square trinomial, we require:

#c = (b/2)^2#

In our example:

#c = (color(blue)(15)/2)^2 = 225/4#

Feb 24, 2017

Answer:

#c=225/4 = 56.25#

Explanation:

Consider the equaton: #x^2-15x+c=0#

This equation has a single root where its discriminant #=0#

When this equation has a single root its factors will be of the form:

#(x-p)(x-p) = 0 -> (x-p)^2 =0#

Hence: The trinomial will be a perfect square when the discriminant of #x^2-15x+c=0# is equal to #0#

I.e. when #15^2-4*1*c =0#

#4c = 225 -> c=225/4 = 56.25#

To test this result consider #c=225/4 = (15/2)^2#

Hence our trinomial is: #x^2-15x+(15/2)^2#

Which factorises to: #(x-15/2)(x-15/2) = (x-15/2)^2# which is a perfect square.