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$$\sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}} = x$$

We have to find the value of $x$.

Taking the terms to other side and squaring is increasing the power of $x$ rapidly, and it becomes unsolvable mess.

I think the answer lies in simplification, but can't do it. Also I have tried taking $\sqrt{2}$ common, but it doesn't help.

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Squaring the equation: \begin{equation} x^2=2+2\sqrt{1-\frac{3}{4}}=2+1=2+1=3 \end{equation} Finally you get $$x=\sqrt{3}$$

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Recognize the quantity under radical sign is a perfect square, it all cancels out.

$$\sqrt{1+\frac{\sqrt{3}}{2}} = \sqrt{1+2\cdot \frac12 \cdot \frac{\sqrt{3}}{2}} = \sqrt { (\cos \pi/3 + \sin \pi/3 )^2} = (\cos \pi/3 + \sin \pi/3 ) $$

Similarly,

$$\sqrt{1-\frac{\sqrt{3}}{2}} = (\cos \pi/3 - \sin \pi/3 ) $$

Adding, $$ \rightarrow \sqrt 3 $$

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  • $\begingroup$ Wow this is nice! +1 $\endgroup$ – Max Payne Mar 21 '16 at 12:37
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Hint: square the surd and note the cross term may be simplified.

Also note that $\sqrt{3}/2 = \cos{(\pi/6)}$ and you may use the double angle formula $1 +\cos{t} = 2 \cos^2{(t/2)}$, etc.

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  • $\begingroup$ This is my way. $(+1)$ $\endgroup$ – lab bhattacharjee Mar 21 '16 at 17:44
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try converting expr inside surd into a square.

$$x = \sqrt{1+\frac{\sqrt{3}}{2}}+\sqrt{1-\frac{\sqrt{3}}{2}}$$ $$ = \sqrt{\frac{4+2\sqrt{3}}{4}}+\sqrt{\frac{4-2\sqrt{3}}{4}} $$

$$ = \sqrt{\frac{1+3+2\sqrt{3}}{2^2}}+\sqrt{\frac{1+3-2\sqrt{3}}{2^2 }} $$

$$ = \sqrt{\left(\frac{1+\sqrt{3}}{2}\right)^2}+\sqrt{\left(\frac{1-\sqrt{3}}{2}\right)^2} $$

$$ = \frac{|1+\sqrt{3}|}{2} + \frac{|1-\sqrt{3}|}{2}$$ $$ = \frac{1+\sqrt{3}}{2} + \frac{\sqrt{3}-1}{2}$$ $$ = \sqrt{3}$$

Edit

My earlier answer was incorrect since i foolishly took square root and forgot about modulus. I have ammended it. It is important to note that: $$\sqrt{x^2} = |x|$$

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  • 1
    $\begingroup$ How can it possibly be 1! Square roots are non-negative and $1+\frac{\sqrt{3}}{2}>1$. It evaluates to $\sqrt{3}$. Your second square root is negative, when it should be positive! $\endgroup$ – almagest Mar 21 '16 at 12:26
  • $\begingroup$ Oh i must have missed something $\endgroup$ – Max Payne Mar 21 '16 at 12:27
  • $\begingroup$ How is it possible you found this solution? I don't understand how you remove the square roots $\endgroup$ – seoanes Mar 21 '16 at 12:27
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    $\begingroup$ $\sqrt{(1-\sqrt 3)^2}\not=1-\sqrt 3$. $\endgroup$ – mathlove Mar 21 '16 at 12:28

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