How do you solve # (x^2-8x+4)/(x^2-2) = 3#?

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Answer 1 · 2016-06-06T14:40:57

First, let's determine the restrictions on x. This can be done by setting the denominator to 0 and solving for x.

#x^2 - 2 = 0#

#x^2 = 2#

#x = +-sqrt(2)#

Therefore, #x != +-sqrt(2)#

Now we can solve:

#x^2 - 8x + 4 = 3(x^2 - 2)#

#x^2 - 8x + 4 = 3x^2 - 6#

#0 = 2x^2 + 8x -10#

#0 = 2x^2 + 10x - 2x - 10#

#0 = 2x(x + 5) - 2(x + 5)#

#0 = (2x - 2)(x + 5)#

#x = 1 and -5#

Checking our solutions in the original equation we find that both work.

Our solution set is therefore #{x = 1, -5; x !=+-sqrt(2)}#

Hopefully this helps!