How do you solve sqrt(x-5) - sqrt(2x-3) = -2?

Oct 14, 2015

Move one of the radicals to the other side of the equation, then square both sides a couple of times. Answer: $x = 14$

Explanation:

$\sqrt{x - 5} = \sqrt{2 x - 3} - 2$

${\left(\sqrt{x - 5}\right)}^{2} = {\left(\sqrt{2 x - 3} - 2\right)}^{2}$

$x - 5 = 2 x + 1 - 4 \sqrt{2 x - 3}$

$- x - 6 = - 4 \sqrt{2 x - 3}$

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

${x}^{2} + 12 x + 36 = 32 x - 48$

${x}^{2} - 20 x + 84 = 0$

$\left(x - 6\right) \left(x - 14\right) = 0$

$x = 6 \mathmr{and} x = 14$

Check for extraneous solutions:

Only x = 14 works in the original equation:

$\sqrt{14 - 5} = \sqrt{2 \cdot 14 - 3} - 2$

$3 = 3$

Hope that helped