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

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

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• 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 $b:= \sqrt{1-x}$.

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

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.

You can solve the square of a sum by writing out the square as a product of two sums and writing out the multiplication for each pair of terms.

$$(1+\sqrt{x-1})^2 =\\ (1+\sqrt{x-1})(1+\sqrt{x-1})=\\ 1\times1 + 1\times\sqrt{x-1}+\sqrt{x-1}\times1+\sqrt{x-1}\times\sqrt{x-1}=\\ 1+\sqrt{x-1}+\sqrt{x-1}+x-1=\\ 2\sqrt{x-1}+x$$