# How do you solve for x in this problem? 4x^2 - 3 = -x?

May 12, 2018

$x = - 1 \text{ or } x = \frac{3}{4}$

#### Explanation:

$\text{rearrange into standard form } \textcolor{w h i t e}{x} a {x}^{2} + b x + c = 0$

$\text{add "x" to both sides}$

$\Rightarrow 4 {x}^{2} + x - 3 = 0 \leftarrow \textcolor{b l u e}{\text{in standard form}}$

$\text{factor the quadratic using the a-c method}$

$\text{the factors of the product } 4 \times - 3 = - 12$

$\text{which sum to + 1 are + 4 and - 3}$

$\text{split the middle term using these factors}$

$4 {x}^{2} + 4 x - 3 x - 3 = 0 \leftarrow \textcolor{b l u e}{\text{factor by grouping}}$

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

$\text{take out the "color(blue)"common factor } \left(x + 1\right)$

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

$\text{equate each factor to zero and solve for x}$

$x + 1 = 0 \Rightarrow x = - 1$

$4 x - 3 = 0 \Rightarrow x = \frac{3}{4}$

May 12, 2018

#### Explanation:

First we rearrange;

$4 {x}^{2} - 3 = - x$
$4 {x}^{2} + x - 3 = 0$

now we have $a {x}^{2} + b x + c = 0$ format, so we can use our quadratic equation to solve.

$x = \frac{- b + \sqrt{{b}^{2} - 4 a c}}{2 a}$
and
$x = \frac{- b - \sqrt{{b}^{2} - 4 a c}}{2 a}$
After subbing everything in we get $x = - 1$ or $x = .75$

May 12, 2018

$x = - 1 \mathmr{and} x = \frac{3}{4}$

#### Explanation:

show below

$4 {x}^{2} - 3 = - x$

$4 {x}^{2} + x - 3 = 0$

$\left(x + 1\right) \cdot \left(4 x - 3\right) = 0$

$x + 1 = 0 \implies x = - 1$

$4 x - 3 = 0 \implies 4 x = 3 \implies x = \frac{3}{4}$