SOLUTION: How to solve x=√3x-12 +4 radical sign over 3x-12

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Question 135978: How to solve x=√3x-12 +4 radical sign over 3x-12

Answer by solver91311(12126) About Me  (Show Source):
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x=sqrt%283x-12%29%2B4

Isolate the radical:
x-4=sqrt%283x-12%29

Square both sides:
x%5E2-8x%2B16=3x-12

Put the quadratic in standard form:
x%5E2-8x-3x%2B16%2B12=0
x%5E2-11x%2B28=0

Since -7%2A-4=28 and -7-4=-11, this quadratic factors, so:
%28x-7%29%28x-4%29=0

Therefore, by the zero product rule:
x-7=0 or x-4=0
x=7 or x=4

Since we had to square both sides of the equation, there is the possibility that we introduced an extraneous root -- one that is a solution to the derived equation but is NOT a solution to the original equation. Check both answers.

Is 7=sqrt%283%2A7-12%29%2B4 a true statement? sqrt%2821-12%29%2B4=sqrt%289%29%2B4=3%2B4=7 Checks, the equation is true when x=7, so x=7 is a valid solution.

Is 4=sqrt%283%2A4-12%29%2B4 a true statement? sqrt%2812-12%29%2B4=sqrt%280%29%2B4=0%2B4=0 Checks, the equation is true when x=4, so x=4 is a valid solution.

No extraneous roots were introduced, so both solutions are valid.