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The value of $x$ (considering only the positive root) satisfying the equation $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \sqrt{\frac{x}{2}}$$ is?

Please, can you guys help me out because I can't understand which formula to use, Do we have to use the Discriminant formula ($b^2-4ac$)?

miracle173
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mohammad yaseen
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    Your title should be descriptive. It should not contain the entirety of the question unless it is short. And your expression is ambiguous. In the first pair of parentheses, you have $(\sqrt{5}+2\sqrt{6})$ and in your third pair of parentheses you have $(\sqrt{5+2\sqrt{6}})$ And your left-facing parenthesis in front of the / has no matching right-facing parenthesis. After the $=$ sign it is unclear if you want $\frac{\sqrt{x}}{2}$ or $\sqrt{\frac{x}{2}}$. – John Wayland Bales Jul 29 '17 at 06:57
  • Here's a MathJax tutorial :) – Shaun Jul 29 '17 at 07:28
  • Edited @John Wayland Bales – mohammad yaseen Jul 29 '17 at 07:51
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    @All: In case it matters, the OP is working on the problem himself. See the comment under N.F.Taussig's answer. Mohammad, to avoid unpleasant surprises in the future you may want to check out our guidelines on asking. – Jyrki Lahtonen Jul 29 '17 at 08:54
  • this close votes are ridiculous – miracle173 Jul 29 '17 at 09:30
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    the right hand side of the equation in the title differs from the body of your text. please make the appropriate changes. – miracle173 Jul 29 '17 at 09:33
  • May help simplifying the nested radicals $\sqrt{5+2 \sqrt{6}}=\sqrt{2}+\sqrt{3}$ and $\sqrt{5-2 \sqrt{6}}=\sqrt{2}-\sqrt{3}$. There is a formula for this$\quad \sqrt {a+{\sqrt {b}}}=\sqrt {\frac {a+{\sqrt {a^{2}-b}}}{2}}+\sqrt {\frac {a-{\sqrt {a^{2}-b}}}{2}}$ and $\quad \sqrt {a-{\sqrt {b}}}=\sqrt {\frac {a+{\sqrt {a^{2}-b}}}{2}}-\sqrt {\frac {a-{\sqrt {a^{2}-b}}}{2}}$ – Raffaele Jul 29 '17 at 14:39

4 Answers4

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Let

$ a = \sqrt{5+ 2\sqrt{6}}, \ \ b= \sqrt{5- 2\sqrt{6}}.$

$$ \frac{ a +b}{a-b} = \sqrt{\frac{x}{2}}$$

$$ \frac{(a+b)(a+b)}{(a-b)(a+b)} = \frac{(a+b)^2}{a^2 - b^2} = \sqrt{\frac{x}{2}}$$

$$ \frac{\left(\sqrt{5+ 2\sqrt{6}} + \sqrt{5-2\sqrt{6}}\right)^2}{\left(\sqrt{5+2\sqrt{6}}\right)^2-\left(\sqrt{5-2\sqrt{6}})\right)^2} = \sqrt{\frac{x}{2}}$$

$$ \frac{5+ 2\sqrt{6}+ 2\sqrt{(5+2\sqrt{6})(5 - 2\sqrt{6})}+5 -2\sqrt{6}}{5 +2\sqrt{6} -5 +2\sqrt{6}} = \sqrt{\frac{x}{2}}$$

$$ \frac{10 +2\sqrt{1}}{4\sqrt{6}} = \sqrt{\frac{x}{2}}$$

$$ \frac{12}{4\sqrt{6}}=\sqrt{\frac{x}{2}}$$

$$ \frac{3}{\sqrt{6}} = \sqrt{\frac{x}{2}}$$

$$ x = 3.$$

JCH
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Hint: Multiply the numerator and denominator of $$\frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} - \sqrt{5 - 2\sqrt{6}}}$$ by the conjugate of the denominator to obtain $$\frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} - \sqrt{5 - 2\sqrt{6}}} \cdot \frac{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}{\sqrt{5 + 2\sqrt{6}} + \sqrt{5 - 2\sqrt{6}}}$$ then simplify.

N. F. Taussig
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    I got root6/2 after rationalizing root(3/2) The the choices I have been given for the problem to find x are 6,3,root3,9/2,SO the answer will be 3 right because for the expression its given root(x/2)? – mohammad yaseen Jul 29 '17 at 08:44
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    Assuming you meant $\sqrt{\frac{x}{2}}$, then the answer is $3$. My apologies for any errors I made in editing the question. – N. F. Taussig Jul 29 '17 at 09:00
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$$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \frac{x}{2}$$

Using componendo dividendo

$$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}+({\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}})}{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}-(\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}})} = \frac{x+2}{x-2}$$

$$\frac{\sqrt{5+2\sqrt{6}} +({\sqrt{5+2\sqrt{6}} })}{\sqrt{5-2\sqrt{6}}+ \sqrt{5-2\sqrt{6}})} = \frac{x+2}{x-2}$$

$$\frac{\sqrt{5+2\sqrt{6}}}{\sqrt{5-2\sqrt{6}}}=\frac{x+2}{x-2}$$

$$\frac{5+2\sqrt{6}}{5-2\sqrt{6}}=(\frac{x+2}{x-2})^2$$

Using it again

$$\frac{5+2\sqrt{6}+(5-2\sqrt{6})}{5+2\sqrt{6}-(5-2\sqrt{6})}=(\frac{(x+2)^2+(x-2)^2}{(x+2)^2-(x-2)^2})$$

$$\frac{10}{4\sqrt6}=\frac{2x^2+8}{8x}$$

$$\frac{10}{4\sqrt6}=\frac{x^2+4}{4x}$$

$$\frac{10}{\sqrt6}=\frac{x^2+4}{x}$$

$$\sqrt6(x^2+4)-10x=0$$

Just apply quadratic formula.

Here you go

Crazy
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Hint: Calculate $$(\sqrt2+\sqrt3)^2.$$ Then do the same with a sign change.

Jyrki Lahtonen
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