# Question #c8554

Feb 16, 2018

The discriminant is $- 31$, meaning there are $2$ complex solutions.

#### Explanation:

The discriminant is found in the quadratic formula:

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

with the discriminant being

${b}^{2} - 4 a c$

Let's call the discriminant $d$:

$d < 0 \to \text{two zeroes that are complex numbers}$

$d = 0 \to \text{one real zero or a repeated zero}$

$d > 0 \to \text{two distinct real zeroes}$

We have the equation ${x}^{2} - 3 x + 10$ already in $a {x}^{2} + b x + c$ standard form, so plug into the formula:

${\left(- 3\right)}^{2} - 4 \cdot 1 \cdot 10$

$9 - 40 = - 31$

$d = - 31$

The discriminant is $- 31$, meaning there are $2$ complex solutions.

Bonus: Finding the complex solutions

$\frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$

$\frac{3 \pm \sqrt{- 31}}{2}$

Assume $\sqrt{- 1}$ is $i$

$\frac{3 \pm \sqrt{31 \cdot - 1}}{2}$

$\frac{3 \pm \sqrt{31} i}{2}$

The zeroes are

$\frac{3 + \sqrt{31} i}{2}$, and $\frac{3 - \sqrt{31} i}{2}$

Here is a graph for reference: graph{x^2-3x+10 [-31.76, 32.96, -1.82, 30.53]}

Have a nice day!

Feb 16, 2018

The discriminant is $- 31$.

#### Explanation:

To find the discriminant, you have to use the quadratic formula:

$\textcolor{w h i t e}{=} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The discriminant part is ${b}^{2} - 4 a c$. We need to identify the $a$, $b$, and $c$ in our quadratic:

$\textcolor{w h i t e}{=} {x}^{2} - 3 x + 10$

$a$ is 1, $b$ is $- 3$, and $c$ is $10$. Now plug these into the discriminant:

$\textcolor{w h i t e}{\implies} {b}^{2} - 4 a c$

$\implies {\left(- 3\right)}^{2} - 4 \left(1\right) \left(10\right)$

$= 9 - 40$

$= - 31$

The discriminant is $- 31$, which means that the quadratic has no real roots.