# Solve $\sqrt{2x-5} - \sqrt{x-1} = 1$

Although this is a simple question I for the life of me can not figure out why one would get a 2 in front of the second square root when expanding. Can someone please explain that to me?

Example: solve $\sqrt{(2x-5)} - \sqrt{(x-1)} = 1$

Isolate one of the square roots: $\sqrt{(2x-5)} = 1 + \sqrt{(x-1)}$

Square both sides: $2x-5 = (1 + \sqrt{(x-1)})^{2}$

We have removed one square root.

Expand right hand side: $2x-5 = 1 + 2\sqrt{(x-1)} + (x-1)$-- I don't understand?

Simplify: $2x-5 = 2\sqrt{(x-1)} + x$

Simplify more: $x-5 = 2\sqrt{(x-1)}$

Now do the "square root" thing again:

Isolate the square root: $\sqrt{(x-1)} = \frac{(x-5)}{2}$

Square both sides: $x-1 = (\frac{(x-5)}{2})^{2}$

Square root removed

Thank you in advance for your help

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## migrated from mathematica.stackexchange.comMay 15 '13 at 10:12

This question came from our site for users of Mathematica.

Welcome here. I'm not entirely sure whether you are on the right place. This is the site for the software Mathematica and not a mathmatics site. – halirutan May 15 '13 at 10:09

I suppose you know this relation: $(a+b)^2=a^2+2ab+b^2$. In the step that you don't understand exactly this relation is used with $a:=1$ and $b:= \sqrt{1-x}$.

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$$2x-5 = (1 + \sqrt{x-1})^2$$

to expand RHS use this formula or simple mulipty it with itself(to do square). formula is:

$(a+b)^2=a^2+b^2+2\times a\times b$

so your expansion will be $$2x-5 = (1^2 + (\sqrt{x-1})^2+2\times1\times \sqrt{x-1})$$ $$2x-5 = (1 + {x-1}+2\times \sqrt{x-1})$$ $$2x-5 = x+2\sqrt{x-1}$$ $$x-5 = 2\sqrt{x-1}$$ now you have your way.

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