SOLUTION: sqrt of 3x + sqrt of 12 = x+5 over sqrt of 3, please help.

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Question 260291: sqrt of 3x + sqrt of 12 = x+5 over sqrt of 3, please help.
Answer by drk(1908) About Me  (Show Source):
You can put this solution on YOUR website!
Here is the original problem:
sqrt%283x%29+%2B+sqrt%2812%29+=+%28x%2B5%29+%2F+sqrt%283%29
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I don't know if the first x is in or outside of the square root. Lets first assume in and see where it takes us.
step 1 - multiply evreything by sqrt(3) to get
sqrt%283%29%2Asqrt%283x%29+%2B+sqrt%2812%29%2Asqrt%283%29+=+x%2B5
step 2 - simplify the square roots to get
3sqrt%28x%29+%2B+6+=+x%2B5
step 3 - subtract 6 from both sides to get
3sqrt%28x%29+=+x-1
step 4 - divide by 3 to get
sqrt%28x%29+=+%281%2F3%29x+-+%281%2F3%29
step 5 - square both sides to get
x+=+%281%2F9%29x%5E2+-%282%2F9%29x+%2B+1%2F9
step 6 - multiply everything by 9 to get
9x+=+x%5E2+-2x+%2B+1
step 7 - set = 0 and factor
0+=+x%5E2+-11x+%2B+1
by quadratic, we get
x+=+%2811+%2B-+sqrt%28+121-4%2A1%2A1+%29%29%2F%282%29+
then
x+=+%2811+%2B-+sqrt%28+117+%29%29%2F%282%29+
and then
x+=+%2811+%2B-+3sqrt%28+13+%29%29%2F%282%29+
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Now, suppose the x is outside the square root. We get
x%2Asqrt%283%29+%2B+sqrt%2812%29+=+%28x%2B5%29+%2F+sqrt%283%29
step 1 - multiply evreything by sqrt(3) to get
x%2Asqrt%283%29%2Asqrt%283%29+%2B+sqrt%2812%29%2Asqrt%283%29+=+x%2B5
step 2 - simplify the square roots to get
3x+%2B+6+=+x%2B5
step 3 - subtract x from both sides to get
2x+%2B+6+=+5
step 4 - subtract 6 from both sides to get
2x+=+-1
divide by 2 to get
x+=+-1%2F2