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

##### 2 Answers
Feb 24, 2017

$c = {\left(\frac{15}{2}\right)}^{2} = \frac{225}{4}$

#### Explanation:

We find:

${\left(x + \frac{b}{2}\right)}^{2} = {x}^{2} + 2 \left(x\right) \left(\frac{b}{2}\right) + {\left(\frac{b}{2}\right)}^{2} = {x}^{2} + b x + {b}^{2} / 4$

So in order for ${x}^{2} + b x + c$ to be a perfect square trinomial, we require:

$c = {\left(\frac{b}{2}\right)}^{2}$

In our example:

$c = {\left(\frac{\textcolor{b l u e}{15}}{2}\right)}^{2} = \frac{225}{4}$

Feb 24, 2017

$c = \frac{225}{4} = 56.25$

#### Explanation:

Consider the equaton: ${x}^{2} - 15 x + c = 0$

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

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

$\left(x - p\right) \left(x - p\right) = 0 \to {\left(x - p\right)}^{2} = 0$

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

I.e. when ${15}^{2} - 4 \cdot 1 \cdot c = 0$

$4 c = 225 \to c = \frac{225}{4} = 56.25$

To test this result consider $c = \frac{225}{4} = {\left(\frac{15}{2}\right)}^{2}$

Hence our trinomial is: ${x}^{2} - 15 x + {\left(\frac{15}{2}\right)}^{2}$

Which factorises to: $\left(x - \frac{15}{2}\right) \left(x - \frac{15}{2}\right) = {\left(x - \frac{15}{2}\right)}^{2}$ which is a perfect square.