We have removed one square root.
Expand right hand side: 2x−5 = 1 + 2√(x−1) + (x−1)
Simplify: 2x−5 = 2√(x−1) + x
Subtract x from both sides: x−5 = 2√(x−1)
Now do the "square root" thing again:
isolate the square root: √(x−1) = (x−5)/2square both sides: x−1 = ((x−5)/2)2We have now successfully removed both square roots.
Let us continue on with the solution.
Expand right hand side: x−1 = (x2 − 10x + 25)/4
It is a Quadratic Equation! So let us put it in standard form.
Multiply by 4 to remove division: 4x−4 = x2 − 10x + 25
Bring all to left: 4x − 4 − x2 + 10x − 25 = 0
Combine like terms: −x2 + 14x − 29 = 0
Swap all signs: x2 − 14x + 29 = 0
Using the Quadratic Formula (a=1, b=−14, c=29) gives the solutions:
2.53 and 11.47 (to 2 decimal places)
Let us check the solutions:
2.53: √(2·2.53−5) − √(2.53−1) ≈ −1 Oops! Should be plus 1!
11.47: √(2·11.47−5) − √(11.47−1) ≈ 1 Yes that one works.
There is really only one solution:
Answer: 11.47 (to 2 decimal places)
