# How do you find the value of the discriminant and state the type of solutions given r^2+5r+2=0?

Apr 27, 2017

See explanation.

#### Explanation:

The discriminant of a quadratic equation can be calculated as:

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

Here we have:

$a = 1$, $b = 5$, $c = 2$

So the discriminant is:

$\Delta = {5}^{2} - 4 \cdot 1 \cdot 2 = 25 - 8 = 17$

To state the type and number of solutions we use the following property of a quadratic equation:

The quadratic equation has:

• no real solutions if $\Delta < 0$
• one real solution if $\Delta = 0$
• two real solutions if $\Delta > 0$

Here the discriminant is positive, so the equation has 2 different real solutions.

Apr 27, 2017

$\Delta = 17$ telling us that this quadratic equation has two distinct irrational roots.

#### Explanation:

Given:

${r}^{2} + 5 r + 2 = 0$

Note that this is of the form:

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

with $a = 1$, $b = 5$ and $c = 2$.

This has discriminant $\Delta$ given by the formula:

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

$\textcolor{w h i t e}{\Delta} = {\textcolor{b l u e}{5}}^{2} - 4 \left(\textcolor{b l u e}{1}\right) \left(\textcolor{b l u e}{2}\right)$

$\textcolor{w h i t e}{\Delta} = 25 - 8$

$\textcolor{w h i t e}{\Delta} = 17$

Since $\Delta > 0$ this quadratic has two distinct real roots, but because $\Delta$ is not a perfect square those roots are irrational.

The possible cases are:

• $\Delta > 0$ with $\Delta$ a perfect square: Two distinct rational roots.

• $\Delta > 0$ with $\Delta$ not a perfect square: Two distinct irrational roots.

• $\Delta = 0$: One repeated rational root.

• $\Delta < 0$: Two distinct non-real Complex roots, which are complex conjugates of one another.

$\textcolor{w h i t e}{}$
Bonus

We can find the roots using the quadratic formula:

$r = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$\textcolor{w h i t e}{r} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{r} = \frac{- 5 \pm \sqrt{17}}{2}$

$\textcolor{w h i t e}{r} = - \frac{5}{2} \pm \frac{\sqrt{17}}{2}$