# How do you find the nature of the roots using the discriminant given 9x^2 + 6x + 1 = 0?

Nov 2, 2017

See a solution process below:

#### Explanation:

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:

$\textcolor{red}{9}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{6}$ for $\textcolor{b l u e}{b}$

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

${\textcolor{b l u e}{6}}^{2} - \left(4 \cdot \textcolor{red}{9} \cdot \textcolor{g r e e n}{1}\right) \implies$

$36 - 36 \implies$

$0$

Because the Discriminant is $0$ there is just ONE solution for this equation.
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