# What is the discriminant of d^2− 7d + 8 = 0 and what does that mean?

##### 1 Answer
Jul 28, 2015

For this quadratic, $\Delta = 17$, which means that the equation has two distinct real roots.

#### Explanation:

For a quadratic equation written in the general form

$a {x}^{2} + b x + c = 0$

the determinant is equal to

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

Your quadratic looks like this

${d}^{2} - 7 d + 8 = 0$,

which means that, in your case,

$\left\{\begin{matrix}a = 1 \\ b = - 7 \\ c = 8\end{matrix}\right.$

The determinant for your equation will thus be equal to

$\Delta = {\left(- 7\right)}^{2} - 4 \cdot \left(1\right) \cdot \left(8\right)$

$\Delta = 49 - 32 = \textcolor{g r e e n}{17}$

When $\Delta > 0$, the quadratic will have two distinct real roots of the general form

${x}_{1 , 2} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

Because the discriminant is not a perfect square, the two roots will be irrational numbers.

In your case, these two roots will be

${d}_{1 , 2} = \frac{- \left(- 7\right) \pm \sqrt{17}}{2 \cdot 1} = \left\{\begin{matrix}{d}_{1} = \frac{7}{2} + \frac{\sqrt{17}}{2} \\ {d}_{2} = \frac{7}{2} - \frac{\sqrt{17}}{2}\end{matrix}\right.$