# How do you solve (5n - 1) (n+1) = 0?

Dec 23, 2016

$n = \frac{1}{5}$ $\mathmr{and} n = - 1$

#### Explanation:

$\left(5 n - 1\right) \left(n + 1\right) = 0$

$5 {n}^{2} + 4 n - 1 = 0$

$\left(5 n - 1\right) = 0$

$5 n = 1$

$n = \frac{1}{5}$

or

$\left(n + 1\right) = 0$

$n = - 1$

substitute $n = \frac{1}{5}$

$5 {\left(\frac{1}{5}\right)}^{2} + 4 \left(\frac{1}{5}\right) - 1 = 0$

$5 \left(\frac{1}{25}\right) + \frac{4}{5} - 1 = 0$

$\frac{5}{25} + \frac{4}{5} - 1 = 0$

$\frac{1}{5} + \frac{4}{5} - 1 = 0$

$\frac{5}{5} - 1 = 0$

$1 - 1 = 0$

substitute $n = - 1$

$5 {n}^{2} + 4 n - 1 = 0$

5(-1)^2+(4(-1)-1=0

$5 + \left(- 4\right) - 1 = 0$

$1 - 1 = 0$

Dec 25, 2016

$n = \frac{1}{5} \text{ or } n = - 1$

#### Explanation:

$\left(5 n - 1\right) \left(n + 1\right) = 0$

If the product of 2 factors is 0, one of them must be 0.

Consider each factor equal to 0.

If $5 n - 1 = 0 \textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots} \text{if } n + 1 = 0$

$\rightarrow \text{ "5n = 1color(white)(..............)rarr" } n = - 1$

$\rightarrow \text{ } n = \frac{1}{5}$