# How do you find the discriminant of 5x^2-4x+1=3x and use it to determine if the equation has one, two real or two imaginary roots?

Oct 2, 2017

See a solution process below:

#### Explanation:

First, we need to put the equation in standard form. Subtract $\textcolor{red}{3 x}$ from each side of the equation to put the equation in standard form while keeping the equation balanced:

$5 {x}^{2} - 4 x - \textcolor{red}{3 x} + 1 = 3 x - \textcolor{red}{3 x}$

$5 {x}^{2} + \left(- 4 - \textcolor{red}{3}\right) x + 1 = 0$

$5 {x}^{2} + \left(- 7\right) x + 1 = 0$

$5 {x}^{2} - 7 x + 1 = 0$

The quadratic formula states:

For $a {x}^{2} + b x + c = 0$, the values of $x$ which are the solutions to the equation are given by:

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

The discriminate is the portion of the quadratic equation within the radical: ${\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}$

If the discriminate is:
- Positive, you will get two real solutions
- Zero you get just ONE solution
- Negative you get complex solutions

To find the discriminant for this problem substitute:

$\textcolor{red}{5}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 7}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{1}$ for $\textcolor{g r e e n}{c}$

${\textcolor{b l u e}{\left(- 7\right)}}^{2} - \left(4 \cdot \textcolor{red}{5} \cdot \textcolor{g r e e n}{1}\right)$

$49 - 20$

$29$

Because the discriminate is positive, you will get two real solutions or roots.

Oct 2, 2017

See below.

#### Explanation:

Arrange $5 {x}^{2} - 4 x + 1 = 3 x$

To get: $5 {x}^{2} - 7 x + 1 = 0$

We now have the form:

$a {x}^{2} + b x + 2$

The discriminant of a quadratic is:

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

if: ${b}^{2} - 4 a c > 0$ then the roots are real and different.

if: ${b}^{2} - 4 a c = 0$ then the roots are real and repeated,( this is sometimes just called a single root).

if: ${b}^{2} - 4 a c < 0$ then the roots are imaginary ( we will have the root of a negative number).

Hope this helps.