# How do use the discriminant to find all values of b for which the equation 2x^2-bx-9=0 has one real root?

Aug 30, 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 value of $b$ where there will be just ONE soluition, we set the discriminate equal to $0$, substitute for $a$ and $c$ and solve for $b$:

Substitute:

$\textcolor{red}{2}$ for $\textcolor{red}{a}$

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

$\textcolor{g r e e n}{- 9}$ for $\textcolor{g r e e n}{c}$

${\textcolor{b l u e}{- b}}^{2} - \left(4 \cdot \textcolor{red}{2} \cdot \textcolor{g r e e n}{- 9}\right) = 0$

${\textcolor{b l u e}{- b}}^{2} - \left(- 72\right) = 0$

${\textcolor{b l u e}{- b}}^{2} + 72 = 0$

${\textcolor{b l u e}{- b}}^{2} + 72 - \textcolor{red}{72} = 0 - \textcolor{red}{72}$

${\textcolor{b l u e}{- b}}^{2} + 0 = - 72$

${\textcolor{b l u e}{- b}}^{2} = - 72$

$\textcolor{red}{- 1} \cdot {\textcolor{b l u e}{- b}}^{2} = \textcolor{red}{- 1} \cdot - 72$

${b}^{2} = 72$

$\sqrt{{b}^{2}} = \pm \sqrt{72}$

$b = \pm \sqrt{36 \cdot 2}$

$b = \pm \sqrt{36} \sqrt{2}$

$b = \pm 6 \sqrt{2}$