# How do you solve sqrt( 7x^2-1) + 1=3x?

Jul 24, 2015

Rewrite this equation as a quadratic and solve for $x$.

#### Explanation:

The first thing you need to do in order to solve this equation is isolate the radical on one side of the equation.

To do that, add $- 1$ to both sides of the equation

sqrt(7x""^2 - 1) + cancel(1) - cancel(1) = 3x - 1

sqrt(7x""^2 - 1) = 3x-1

Next, square both sides of the equation to get rid of the radical

(sqrt(7x""^2 - 1))^2 = (3x-1)^2

$7 x {\text{^2 - 1 = 9x}}^{2} - 6 x + 1$

Move all the terms on one side of the equation

2x""^2 -6x +2 = 0

You can use the quadratic formula to determine the two solutions

${x}_{1 , 2} = \frac{6 \pm \sqrt{36 - 4 \cdot 2 \cdot 2}}{4}$

${x}_{1 , 2} = \frac{6 \pm \sqrt{20}}{4} = \frac{3 \pm \sqrt{5}}{2} \implies \left\{\begin{matrix}{x}_{1} = \frac{3 + \sqrt{5}}{2} \\ {x}_{2} = \frac{3 - \sqrt{5}}{2}\end{matrix}\right.$

You need to check these two solutions to see if both are valid, i.e. if you don't have an extraneous solution.

For ${x}_{1}$ you have

$\sqrt{7 \cdot {\left(\frac{3 + \sqrt{5}}{2}\right)}^{2} - 1} = 3 \cdot \left(\frac{3 + \sqrt{5}}{2}\right) - 1$

This is equivalent to

$\frac{\sqrt{94 + 42 \sqrt{5}}}{\cancel{2}} = \frac{7 + 3 \sqrt{5}}{\cancel{2}}$

Square both sides of the equation to get

${\left(\sqrt{94 + 42 \sqrt{5}}\right)}^{2} = {\left(7 + 3 \sqrt{5}\right)}^{2}$

$94 + 42 \sqrt{5} = 49 + 42 \sqrt{5} + 45$ $\to$ ${x}_{1}$ is $\textcolor{g r e e n}{\text{valid}}$

For ${x}_{2}$ you have

$\sqrt{7 \cdot {\left(\frac{3 - \sqrt{5}}{2}\right)}^{2} - 1} = 3 \cdot \left(\frac{3 - \sqrt{5}}{2}\right) - 1$

This is equivalent to

94 - 42sqrt(5)) = 49 - 42sqrt(5) + 45 $\to$ ${x}_{2}$ is $\textcolor{g r e e n}{\text{valid}}$