How do you solve sqrt(2x-4) = x-2 and find any extraneous solutions?

Aug 27, 2016

Thus, the Soln. is $x = 2 , \mathmr{and} , x = 4$

The eqn. has no extraneous soln.

Explanation:

$\sqrt{2 x - 4} = x - 2$.

$\therefore {\left\{\sqrt{2 x - 4}\right\}}^{2} = {\left(x - 2\right)}^{2}$.

$\therefore 2 \left(x - 2\right) = {\left(x - 2\right)}^{2}$.

$\therefore 2 \left(x - 2\right) - {\left(x - 2\right)}^{2} = 0$

$\therefore \left(x - 2\right) \left(2 - x + 2\right) = 0$

$\therefore \left(x - 2\right) \left(4 - x\right) = 0$

$\therefore x = 2 , \mathmr{and} , x = 4$.

These roots satisfy the given Eqn.

Thus, the Soln. is $x = 2 , \mathmr{and} , x = 4$

The eqn. has no extraneous soln.