SOLUTION: square root of 2x-1 (-) the square root of x-5 = 3

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Question 53925: square root of 2x-1 (-) the square root of x-5 = 3
Answer by funmath(2933) About Me  (Show Source):
You can put this solution on YOUR website!
sqrt%282x-1%29-sqrt%28x-5%29=3
sqrt%282x-1%29-sqrt%28x-5%29%2Bsqrt%28x-5%29=sqrt%28x-5%29%2B3
sqrt%282x-1%29=sqrt%28x-5%29%2B3
%28sqrt%282x-1%29%29%5E2=%28sqrt%28x-5%29%2B3%29%5E2
2x-1=%28sqrt%28x-5%29%2B3%29%28sqrt%28x-5%29%2B3%29
2x-1=%28sqrt%28x-5%29%29%5E2%2B3sqrt%28x-5%29%2B3sqrt%28x-5%29%2B9
2x-1=x-5%2B6sqrt%28x-5%29%2B9
2x-1=x%2B4%2B6sqrt%28x-5%29
2x-x-1-4=x-x%2B4-4%2B6sqrt%28x-5%29
x-5=6sqrt%28x-5%29
%28x-5%29%5E2=%286sqrt%28x-5%29%29%5E2
%28x-5%29%28x-5%29=%286%29%5E2%2A%28sqrt%28x-5%29%29%5E2
x%5E2-5x-5x%2B25=36%28x-5%29
x%5E2-10x%2B25=36x-180
x%5E2-10x-36x%2B25%2B180=36x-36x-180%2B180
x%5E2-46x%2B205=0
%28x-41%29%28x-5%29=0
x-41=0
x-41+41=0+41
x=41
x-5=0
x-5+5=0+5
x=5
It appear that the solutions are x={5,41}. However, you have to be especially careful to check problems that involve even roots for extraneous (false) solutions.
Substitute 5 back into the original equation:
sqrt%282%285%29-1%29-sqrt%28%285%29-5%29=3
sqrt%2810-1%29-sqrt%285-5%29=3
sqrt%289%29-sqrt%280%29=3
3-0=3
3=3 (x=5 checks out)
Substitute 41 back into the original equation:
sqrt%282%2841%29-1%29-sqrt%28%2841%29-5%29=3
sqrt%2882-1%29-sqrt%2841-5%29=3
sqrt%2881%29-sqrt%2836%29=3
9-6=3
3=3 (x=41 checks out)
We can now safely say that x={5,41}
Happy Calculating!!!!