Hi there! Hopefully this helps!
Answer: 11.47(to 2 decimal places).
Isolate one of the square roots: √(2x−5) = 1 + √(x−1)
Square both sides: 2x−5 = (1 + √(x−1))^2
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)/2
Square both sides: x−1 = ((x−5)/2)^2
<em>We have now successfully removed both square roots.</em>
<em>Let's continue with the solution.</em>
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Expand right hand side: x−1 = (x^2 − 10x + 25)/4
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<em>Since it is a Quadratic Equation! let's put it in standard form.</em>
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Multiply by 4 to remove division: 4x−4 = x^2 − 10x + 25
Bring all to left: 4x − 4 − x^2 + 10x − 25 = 0
Combine like terms: −x^2 + 14x − 29 = 0
Swap all signs: x^2 − 14x + 29 = 0
<em>Using the Quadratic Formula (a=1, b=−14, c=29) gives the solutions:</em>
<u><em>2.53</em></u><em> and </em><u><em>11.47 </em></u><em>(to 2 decimal places)</em>
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<em>2.53: √(2×2.53−5) − √(2.53−1) ≈ −1 Oops! Should be plus 1. So it is not the solution.</em>
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<em>11.47: √(2×11.47−5) − √(11.47−1) ≈ 1 Yes that one works. </em>
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