How do you solve #sqrt(6x - 8) = x #?

1 Answer
Oct 2, 2017

Answer:

#x = 2" "# or #" "x = 4#

Explanation:

Given:

#sqrt(6x-8) = x#

Note that #sqrt(...) >= 0#. So we require #x >= 0#

Square both sides of the given equation (noting that this may introduce extraneous solutions), to get:

#6x-8 = x^2#

Subtract #6x-8# from both sides to get:

#0 = x^2-6x+8 = (x-4)(x-2)#

So:

#x = 2" "# or #" "x = 4#

Since these are both positive, they are both solutions of the original equation.