How do you solve sqrt[x+2]-sqrt[6x]=-1 and find any extraneous solutions?

Feb 20, 2018

$\sqrt{x + 2} = \sqrt{6 x} - 1$

${\left(\sqrt{x + 2}\right)}^{2} = {\left(\sqrt{6 x} - 1\right)}^{2}$

$x + 2 = 6 x - 2 \sqrt{6 x} + 1$

$- 5 x + 1 = - 2 \sqrt{6 x}$

${\left(\frac{- 5 x + 1}{-} 2\right)}^{2} = {\left(\sqrt{6 x}\right)}^{2}$

$\frac{25 {x}^{2} - 10 x + 1}{4} = 6 x$

$\left(25 {x}^{2} - 10 x + 1\right) = 24 x$

$25 {x}^{2} - 34 x + 1 = 0$

$x = \frac{34 \left(\frac{+}{-}\right) \sqrt{{34}^{2} - 4 \cdot 25 \cdot 1}}{2 \cdot 25}$

$x = 0.03 , 1.33$ and 0.03 does not work when you plug it back into the equation to solve for x, so it is extraneous