# How do you solve (2x-1)(x+3)=0?

Mar 16, 2016

$x = \frac{1}{2} \text{ or } x = - 3$

#### Explanation:

For this equation to become the value of zero it means that either the left bracket is zero or the right bracket is zero.

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Consider $\textcolor{b l u e}{\left(2 x - 1\right) = 0}$

$\textcolor{m a \ge n t a}{\text{Detailed calculation to show how it works}}$

Add $\textcolor{red}{1}$ to both sides

$\text{ } \textcolor{b l u e}{2 x - 1 \textcolor{red}{+ 1} = 0 \textcolor{red}{+ 1}}$

$\text{ } \textcolor{b l u e}{2 x + 0 = 1}$

Divide both sides by $\textcolor{red}{2}$

" "color(blue)(2/(color(red)(2))xx x=1/(color(red)(2))

But $\frac{2}{2} = 1$ giving

$x = \frac{1}{2}$

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Consider $\textcolor{b l u e}{\left(x + 3\right) = 0}$

$\textcolor{m a \ge n t a}{\text{Calculation skipping steps}}$

$x = - 3$
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Mar 16, 2016

$x$ can equal $- \frac{1}{2}$ or $- 3$

#### Explanation:

$\left(2 x - 1\right) \left(x + 3\right) = 0$

To solve for the value of $x$ simply take each of the binomial factors and set them equal to zero.

$2 x + 1 = 0$
$2 x \cancel{+ 1} \cancel{-} 1 = 0 - 1$
$\frac{\cancel{2} x}{\cancel{2}} = - \frac{1}{2}$
$x = - \frac{1}{2}$

$x + 3 = 0$
$x \cancel{+ 3} \cancel{-} 3 = 0 - 3$
$x = - 3$

$x$ can equal $- \frac{1}{2}$ or $- 3$