# How do you solve 4x - (x * 3^(1/2)) = 6?

Oct 13, 2015

$x = \frac{6}{13} \cdot \left(4 + \sqrt{3}\right)$

#### Explanation:

The idea hee is to isolate $x$ on one side of the equation.

To do that, start by using $x$ as a common factor for the expression that's on the left-hand side of the equation

$4 x - x \cdot {3}^{\frac{1}{2}} = 6$

$x \cdot \left(4 - {3}^{\frac{1}{2}}\right) = 6$

You can rewrite the equation by replacing the fractional exponent by its corresponding radical term

$x \cdot \left(4 - \sqrt{3}\right) = 6$

DIvide both sides of the equation by $\left(4 - \sqrt{3}\right)$ to isolate $x$ on the left-hand side of the equation

$\frac{x \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{4 - \sqrt{3}}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{4 - \sqrt{3}}}}} = \frac{6}{4 - \sqrt{3}}$

$x = \frac{6}{4 - \sqrt{3}}$

Rationalize the denominator of the fraction by multiplying it by $1 = \frac{4 + \sqrt{3}}{4 + \sqrt{3}}$. The expression $\left(4 + \sqrt{3}\right)$ is the conjugate of $\left(4 - \sqrt{3}\right)$. The fraction will thus be equivalent to

6/(4 - sqrt(3)) * (4 + sqrt(3))/(4 + sqrt(3)) = (6 * (4 + sqrt(3)))/((4 - sqrt(3))(4 + sqrt(3))

$= \frac{6 \cdot \left(4 + \sqrt{3}\right)}{{4}^{2} - {\left(\sqrt{3}\right)}^{2}}$

$= \frac{6}{13} \cdot \left(4 + \sqrt{3}\right)$

Therefore, $x$ will be equal to

$x = \textcolor{g r e e n}{\frac{6}{13} \cdot \left(4 + \sqrt{3}\right)}$