SOLUTION: {{{x=sqrt(x-1)+3}}}

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Question 742096: x=sqrt%28x-1%29%2B3
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
We can isolate the square root on one side of the equal sign.
After that we can square both sides of the equal sign to get an equation without a square root.
The new equation will have the same solutions as the original equation, and could have some extra solutions, that we call "extraneous solutions".
We do not worry about extraneous solutions at the beginning, because we will check all solutions. We will check to see if they are really solutions to the original equation. We can eliminate any and all extraneous solutions with that checking step.

FINDING POSSIBLE SOLUTIONS:
x=sqrt%28x-1%29%2B3 --> x-3=sqrt%28x-1%29 --> %28x-3%29%5E2=%28sqrt%28x-1%29%29%5E2 --> %28x-3%29%5E2=x-1 --> x%5E2-6x%2B9=x-1 --> x%5E2-7x%2B10=0
Factoring,
x%5E2-7x%2B10=0 --> %28x-2%29%28x-5%29=0
(You could also solve x%5E2-7x%2B10=0 by applying the quadratic formula or by completing the square, if you do not like factoring)
Any way we solve that quadratic equation, we get the tentative solutions x=2 or x=5
They are solutions of %28x-3%29%5E2=x-1,
but are they solutions of x=sqrt%28x-1%29%2B3?

CHECKING:
For x=2, sqrt%28x-1%29%2B3=sqrt%282-1%29%2B3=1%2B3=4%3C%3E2 shows that x=2 is not a solution of x=sqrt%28x-1%29%2B3.
It is an extraneous solution.

The value x=5 results in sqrt%28x-1%29%2B3=sqrt%285-1%29%2B3=sqrt%284%29%2B3=2%2B3=5, which makes x=sqrt%28x-1%29%2B3 true.
That means that highlight%28x=5%29 is truly a solution of x=sqrt%28x-1%29%2B3. It is the solution to x=sqrt%28x-1%29%2B3.