Question

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

Answer 1

`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`

Answer 2

`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.