# How do you solve sqrt(5x-9)-1=sqrt( 3x-6) and find any extraneous solutions?

Jul 6, 2018

$x = 5$ or $x = 2$

#### Explanation:

We have $x \setminus \ge q 2$ since the radicand must be non negative.
Writing your equation in the form
$\sqrt{5 x - 9} = \sqrt{3 x - 6} + 1$

squaring we get

$5 x - 9 = 3 x - 6 + 1 + 2 \sqrt{3 x - 6}$
collecting like Terms and isolating the square root

$x - 2 = \sqrt{3 x - 6}$
since $x \setminus \ge q 2$ we can square again

${x}^{2} - 7 x + 10 = 0$

${x}_{1 , 2} = \frac{7}{2} \pm \sqrt{\frac{49}{4} - \frac{40}{4}}$

so

${x}_{1} = 5$

${x}_{2} = 2$
checking both Solutions we get that both fulfill the equation above.

Jul 6, 2018

${x}_{1} = 2$ and ${x}_{2} = 5$

#### Explanation:

$\sqrt{5 x - 9} - 1 = \sqrt{3 x - 6}$

$5 x - 9 + 1 - 2 \sqrt{5 x - 9} = 3 x - 6$

$2 x - 2 = 2 \sqrt{5 x - 9}$

$x - 1 = \sqrt{5 x - 9}$

${\left(x - 1\right)}^{2} = 5 x - 9$

${x}^{2} - 2 x + 1 = 5 x - 9$

${x}^{2} - 7 x + 10 = 0$

$\left(x - 2\right) \cdot \left(x - 5\right) = 0$

So, ${x}_{1} = 2$ and ${x}_{2} = 5$ are solutions of this equation.