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I have a question on how the simplification process goes for squaring square-roots and how absolute values interact with one another. Question is from IMO Math-Olympiad 1959 #2

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

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

???

$A^2 = 2(x + |x-1|)$

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Nick
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    Amend the first line of your question ? \begin{eqnarray} A = \sqrt{x - \sqrt{2x - 1}}+ \color{red}{\sqrt{x +\sqrt{2x - 1}}} \end{eqnarray} – Donald Splutterwit Oct 17 '18 at 23:37
  • Wait. You were told those two things were true? So $A^2 = x -\sqrt{2x -1} = x + \sqrt{2x -1} + 2(\sqrt{x + \sqrt{2x -1}})(\sqrt{x - \sqrt{2x -1}}) + x - \sqrt{2x -1}$???????? – fleablood Oct 18 '18 at 00:17
  • My mistake, I have edited and fixed it. Thank you – Nick Oct 18 '18 at 00:47

1 Answers1

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By definition, a square root is always positive (or $0$). So $\sqrt{x^2}=|x|\geq0$ . For example, $\sqrt{(-3)^2}=\sqrt9=3=|{-3}|$.

We also have, for positive $x$ and $y$ , $\sqrt{xy}=\sqrt x\sqrt y$ , and $\sqrt x^2=x=|x|$.

(If they're negative, the expressions are undefined, unless you use imaginary numbers. Then there's the problem of $1=\sqrt1=\sqrt{-1\times-1}\neq\sqrt{-1}\times\sqrt{-1}=i\times i=-1$ .)

Thus (assuming that everything in a $\sqrt\;$ is non-negative, which is true if $x\geq1/2$), the middle expression in $A^2$ is

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

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

(use the difference of squares $(a-b)(a+b)=a^2-b^2$)

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

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

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

$$=2|x-1|$$

mr_e_man
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