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Question: Simplify $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}} + \frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$

My Attempt: On rationalizing both fractions separately and then adding (since denominator becomes $\sqrt{3}$) I got $$-2\sqrt{6}+2\sqrt{2+\sqrt{3}}+2\sqrt{2-\sqrt{3}}$$ However the given answer is $$\frac{\sqrt{6}}{3}$$ Even more confusing is that on inputting the problem into wolframalpha the solution is given as $$\sqrt{2}$$ I have broken my head over this for a couple of hours and I just can't find a solution. Hope someone can help.

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  • $\begingroup$ Also note that inside the root $\sqrt{2\pm\sqrt3}$, $$2\pm \sqrt 3 = \left(\frac32 + \frac 12\right) \pm 2\sqrt{\frac3{2^2}} = \left(\sqrt{\frac{3}{2}}\pm\sqrt{\frac{1}{2}}\right)^2$$ $\endgroup$
    – peterwhy
    Jul 30, 2021 at 12:02

2 Answers 2

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You are right, the correct result for the sum of the two given ratios is $\sqrt{2}$. It is $\frac{\sqrt{6}}{3}$ when you take the difference. Presumably there is a typo in your book.

Revise your work and notice that $$\left(\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\right)^2 =2+\sqrt{3}+2\underbrace{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}_1+2-\sqrt{3}=6,$$ and $$\left(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\right)^2 =2+\sqrt{3}-2\underbrace{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}_1+2-\sqrt{3}=2.$$

P.S. As a complement you may take a look at nested radicals. The above computations imply that $$2\sqrt{2\pm\sqrt{3}}=\sqrt{6}\pm\sqrt{2}.$$

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  • $\begingroup$ Ah I see but then I guess I made a mistake somewhere as then my answer would be 0. Thanks a lot! $\endgroup$
    – Anili
    Jul 30, 2021 at 11:45
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    $\begingroup$ @Anili Have you revised your work? Now you should be able to evaluate both the sun and the difference of the two ratios $\endgroup$
    – Robert Z
    Jul 30, 2021 at 12:03
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    $\begingroup$ Yes I was able to solve it thanks a lot! $\endgroup$
    – Anili
    Jul 30, 2021 at 12:07
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Hint

$$2-\sqrt3=\dfrac{(\sqrt3-1)^2}2$$

$$\implies\sqrt2-\sqrt{2-\sqrt3}=\dfrac{2-(\sqrt3-1)}{\sqrt2}$$

$$\implies\dfrac{2-\sqrt3}{\sqrt2-\sqrt{2-\sqrt3}}=\sqrt{\dfrac23}\cdot\dfrac{2-\sqrt3}{\sqrt3-1}$$

$$\dfrac{2+\sqrt3}{\sqrt2+\sqrt{2+\sqrt3}}=?$$

Now take the sum

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  • $\begingroup$ This is an interesting way to solve it! Thanks! $\endgroup$
    – Anili
    Jul 30, 2021 at 12:07

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