SOLUTION: Solve {{{sqrt(x-5)}}}=cube root(x-3)

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Question 622589: Solve sqrt%28x-5%29=cube root(x-3)
Found 2 solutions by jsmallt9, Alan3354:
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
If there is no error in what you posted, then the best I can do for you is to show you how to eliminate the radicals. The problem you posted is not easy to solve. If there is an error in what you posted then maybe what you see below will help you figure out how to do the problem on your own.

sqrt%28x-5%29=root%283%2C+x-3%29
This equation has two kinds of roots, square and cube. TO eliminate the radicals we can either do them one at a time or both together. To eliminate them both with one operation it will help to rewrite them with fractional exponents instead of radicals:
%28x-5%29%5E%281%2F2%29=%28x-3%29%5E%281%2F3%29
Now we raise both sides to the lowest common denominator (LCM) power. (You'll see why shortly.) The LCM of 2 and 3 is 6:
%28%28x-5%29%5E%281%2F2%29%29%5E6=%28%28x-3%29%5E%281%2F3%29%29%5E6
The rule for exponents when raising a power to a power is to multiply the exponents. So this simplifies to:
%28x-5%29%5E3=%28x-3%29%5E2
Now that the radicals are gone, the equation becomes solvable (in theory). As usual we start the solution by simplifying. Cubing the left side, (x-5)(x-5)(x-5), and squaring the right side, (x-3)(x-3), we get:
x%5E3-15x%5E2%2B75x-125=x%5E2-6x%2B9
With these exponents this is not a linear equation. So we want one side to be zero. Subtracting the entire right side from both sides of the equation we get:
x%5E3-16x%5E2%2B81x-134=0
Next we would factor. Then we would use the Zero Product Property to solve the equation. But this does not factor. At least one solution must be irrational.

Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
Solve sqrt%28x-5%29=cube root(x-3)
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Using the cubic the other tutor made,
there's on real solution.
Using a graph,
x =~ 7.875129794162779
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The other 2 roots are complex numbers.