Writing forms of an equation

Question

Let $x>1/2$. What is the simplest form of the expression $(1+\sqrt{2x-1})/(\sqrt{x+\sqrt{2x-1}})$

Let $a=\sqrt{2x-1}$

$(1+a)/(\sqrt{x+a})$

=$(1+a)/(x+a)^{1/2}$

=$(1+a)(\sqrt{x+a})$

=$\sqrt{x+a}*a*\sqrt{x+a}$

=$(x+a)(a)$

=$xa+a^2$

=$(x+a)*a$

Plug in $a=\sqrt{2x-1}$

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

=$x\sqrt{2x-1}+2x-1$

Now plug in $x>1/2$

=$1/2\sqrt{2(1/2)-1}+2(1/2)-1$=$0$

I was trying to solve this problem and was not sure if I was doing it right. I am still not sure on what to do. Can someone please help me with this solution?

I just found out there was an error to the problem posted earlier in the book I was reading from. This is suppose to be the real one.

Answer 1

If you multiply top and bottom by $\sqrt{x-\sqrt{2x-1}}$ the denominator becomes $$ \sqrt{x^2 - (2x - 1)} = \sqrt{(x-1)^2} = x-1 $$

now you've just got to figure out the numerator $$ \left( \sqrt{x-\sqrt{2x-1}} \right) \left( 1+\sqrt{2x+1} \right) $$

Edit: Now that you've fixed the numerator to $(1+\sqrt{2x-1})$ that makes things quite a bit simpler. The other solution is nicer with that fix, though I did figure out that you can get the same answer from here as well. $$ \left( \sqrt{x-\sqrt{2x-1}} \right) \left( 1+\sqrt{2x-1} \right) = \sqrt{(x-\sqrt{2x-1})(1+\sqrt{2x-1})^2} $$ Which ends up simplifying to something similar to above $$ \sqrt{2(x-\sqrt{2x-1})(x+\sqrt{2x-1})} = \sqrt{2(x-1)^2} $$ $$ \frac{\sqrt{2}(x-1)}{(x-1)} = \sqrt{2} $$

Answer 2

$$\begin{align} \text{numerator}^2 = \left(\; 1 + \sqrt{2x-1} \;\right)^2 &= 1^2 \;+\; \sqrt{2x-1}^2 \;+\; 2\cdot 1 \cdot \sqrt{2x-1} \\[4pt] &= 1 + (2x-1) + 2\sqrt{2x-1} \\[4pt] &= 2\;\left(\;x + \sqrt{2x-1}\;\right) \\[4pt] &= 2 \; \text{denominator}^2 \end{align}$$ Since the numerator and denominator are necessarily positive, we can take square roots without sign ambiguity, and then write $$\frac{\text{numerator}}{\text{denominator}} = \frac{\sqrt{2} \;\; \text{denominator}}{\text{denominator}} = \sqrt{2}$$