# How do you solve sqrt[7x+2]-2=7x and find any extraneous solutions?

Jun 15, 2018

$x = - \frac{1}{7}$ or $x = - \frac{2}{7}$

#### Explanation:

Move the constant term

$\sqrt{7 x + 2} = 7 x + 2$

Square both sides

${\left(\sqrt{7 x + 2}\right)}^{2} = {\left(7 x + 2\right)}^{2}$

$7 x + 2 = 49 {x}^{2} + 28 x + 4$

Move everything over

$0 = 49 {x}^{2} + 21 x + 2$

x_(1,2) = (-b ± sqrt(b ^2 - 4ac)) / (2a)

So

x_(1, 2) = (-21 ± sqrt(21 ^2 - 4(49)(2))) / (2(49))

Solve to get

$x = - \frac{1}{7} \text{ }$ or $\text{ } x = - \frac{2}{7}$

To check for extraneous roots, plug the solutions back into the original equation.

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

$1 = 1 \to$ the solution $\left(- \frac{1}{7}\right)$ is correct!

Repeat for the other solution

$\sqrt{7 \left(- \frac{2}{7}\right) + 2} = 7 \left(- \frac{2}{7}\right) + 2$

$0 = 0 \to$ the solution $\left(- \frac{2}{7}\right)$ is also correct!

Extraneous solutions appear when your solutions don't satisfy the original equation.