SOLUTION: I have an unsimplified equation and need to simplify it then solve for x: {{{( 3sqrt(2) - x )/( sqrt(x) ) = sqrt(2x)/ ( sqrt(2) - x ) }}} so if I simplify using cross multipl

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: I have an unsimplified equation and need to simplify it then solve for x: {{{( 3sqrt(2) - x )/( sqrt(x) ) = sqrt(2x)/ ( sqrt(2) - x ) }}} so if I simplify using cross multipl      Log On


   

Question 976853: I have an unsimplified equation and need to simplify it then solve for x:
%28+3sqrt%282%29+-+x+%29%2F%28+sqrt%28x%29+%29+=+sqrt%282x%29%2F+%28+sqrt%282%29+-+x+%29+
so if I simplify using cross multiply I should have?:
6+-+sqrt%282%29x+-+3sqrt%282%29x+%2B+x%5E2+=+sqrt%282%29x
x%5E2+-+5sqrt%282%29x+%2B+6+=+0
I'm new to doing this with square roots all involved, did I go about this right?
Found 2 solutions by josgarithmetic, Alan3354:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
I only partially checked your first step. Assuming your work is correct to the last equation shown, you now have a quadratic equation and can use general solution of a quadratic equation to solve for x.

x%5E2-5sqrt%282%29x%2B6=0

x=%285sqrt%282%29%2B-+sqrt%2825%2A2-4%2A6%29%29%2F2-----you can simplify this yourself, and check if it is acceptable in the originally-given equation.

highlight%28%285sqrt%282%29%2B-+sqrt%2826%29%29%2F2%29

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I have an unsimplified equation and need to simplify it then solve for x:
%28+3sqrt%282%29+-+x+%29%2F%28+sqrt%28x%29+%29+=+sqrt%282x%29%2F+%28+sqrt%282%29+-+x+%29+
so if I simplify using cross multiply I should have?:
6+-+sqrt%282%29x+-+3sqrt%282%29x+%2B+x%5E2+=+sqrt%282%29x
x%5E2+-+5sqrt%282%29x+%2B+6+=+0
I'm new to doing this with square roots all involved, did I go about this right?
================
Looks right. Solve for x.
x%5E2+-+5sqrt%282%29x+%2B+6+=+0
x%5E2+%2B+6+=+5sqrt%282%29x
Square both sides.
x%5E4+%2B+12x%5E2+%2B+36+=+50x%5E2
x%5E4+-+38x%5E2+%2B+36+=+0
---
Sub u for x^2
u%5E2+-+38u+%2B+36+=+0
Solved by pluggable solver: SOLVE quadratic equation (work shown, graph etc)
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-38x%2B36+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-38%29%5E2-4%2A1%2A36=1300.

Discriminant d=1300 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--38%2B-sqrt%28+1300+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%28-38%29%2Bsqrt%28+1300+%29%29%2F2%5C1+=+37.0277563773199
x%5B2%5D+=+%28-%28-38%29-sqrt%28+1300+%29%29%2F2%5C1+=+0.972243622680054

Quadratic expression 1x%5E2%2B-38x%2B36 can be factored:
1x%5E2%2B-38x%2B36+=+%28x-37.0277563773199%29%2A%28x-0.972243622680054%29
Again, the answer is: 37.0277563773199, 0.972243622680054. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-38%2Ax%2B36+%29

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x^2 = answers shown
Find the sq roots.
Check for extraneous solutions.