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

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Answer 1 · 2017-10-02T20:48:28

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

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.