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

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Answer 1 · 2017-02-24T00:03:13

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

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 · 2017-02-24T00:24:44

$c=225/4 = 56.25$

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.

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