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My exams are approaching fast and I found this question in one of the unsolved sample papers. I tried squaring the whole term but couldn't work out the answer.

I am a ninth grader so please try to explain in simple terms.

3 Answers3

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HINT:

What is $\displaystyle (\sqrt{2} + \sqrt{3})^2 $ and $\displaystyle (\sqrt{5} - \sqrt{3})^2 $?


The basic 'trick', so to say, behind such questions is to identify that the surd can be expressed as a square. For example, consider your first surd $\sqrt{5+2\sqrt{6}}$. Here, there's a $2\cdot\sqrt{\text{something}}$. Now, if you see, the factors of that $\sqrt{\text{something}}$ are $\sqrt2$ and $\sqrt3$. A quick check shows that the sum of their squares give $5$. So, this is of the form $a^2 + b^2 + 2ab = (a+b)^2$

Parth Thakkar
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  • Thanks a lot , I didn't think in that direction . Can we simplify this answer more after we get $(\sqrt{2} + \sqrt{3})^2 + (\sqrt{5} - \sqrt{3})^2$ – MayankJain Sep 14 '13 at 09:44
  • Actually you get $\sqrt{(\sqrt 2 + \sqrt 3)^2} + \sqrt{(\sqrt 5 - \sqrt 3)^2}$. – Tunococ Sep 14 '13 at 09:50
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    I like this answer! It is pitched at exactly the right level. – TonyK Sep 14 '13 at 10:38
  • Can't it be $(\sqrt{3} - \sqrt{5})^2$ also ? – MayankJain Sep 14 '13 at 10:46
  • @Mayank, yes it can be. But then we are to consider the square root of that expression (as Tunacoc pointed out). Now, the value of the square root of something is always positive. So, if you write the expression as $ \sqrt{ (\sqrt3 - \sqrt5)^2 } $, then it's alright, but then when you cancel off the $\sqrt{}$ and the square, you've gotta be careful that you don't write $ \sqrt{ (\sqrt3 - \sqrt5)^2 } = \sqrt3 - \sqrt5$ which is negative. Does that remove all doubts? – Parth Thakkar Sep 15 '13 at 05:43
  • @Mayank, I think there's some confusion over here. It should've been clear from the previous comment of mine (and from what Tunacoc said), however, to be sure you don't get it wrong, one thing to be said is: one of the given numbers is $ \sqrt{5+2\sqrt{6}} $. Now, $ (\sqrt{2} + \sqrt{3})^2 \not = \sqrt{5+2\sqrt{6}}$. Just check your calculations. – Parth Thakkar Sep 15 '13 at 05:53
  • Ok , Thanks a lot everyone . – MayankJain Sep 15 '13 at 14:47
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There is a relevant discussion in Dummit and Foote with respect to biquadratic extensions: $\sqrt{a+\sqrt{b}}=\sqrt{m}+\sqrt{n}$ if and only if $a^2-b$ is a perfect square.

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It's worth noting that $(\sqrt a \pm \sqrt b)^2 = (a+b) \pm 2\sqrt{ab}$.

So, if the answer to $\sqrt{M \pm 2\sqrt N}$ is going to be of the form $ \sqrt a \pm \sqrt b$ then we need $N=ab$ and $M = a+b$.

$\sqrt{5+2\sqrt{6}}$

Note that $6 = 2 \times 3$ and $5 = 2 + 3$.

Hence $$(\sqrt 3 + \sqrt 2)^2 = 5+2 \sqrt 6$$


$\sqrt{8-2\sqrt{15}}$

Note that $15 = 5 \times 3$ and $8 = 5 + 3$.

Hence $$(\sqrt 5 - \sqrt 3)^2 = 8 - 2 \sqrt{15}$$

\begin{align} \sqrt{5+2\sqrt{6}} + \sqrt{8-2\sqrt{15}} &= (\sqrt 3 + \sqrt 2) + (\sqrt 5 - \sqrt 3) \\ &= \sqrt 5 + \sqrt 2 \end{align}


Added 1/18/2021

Another way of looking at $\sqrt{M + 2\sqrt N} = \sqrt a + \sqrt b$.

\begin{align} \sqrt{M + 2\sqrt N} &= \sqrt a + \sqrt b \\ M + 2 \sqrt N &= (a+b) + 2\sqrt{ab} \\ \hline M &= a + b \\ N &= ab \\ \hline x^2 - Mx + N &= (x-a)(x-b) \\ \{a,b\} &= \dfrac{M \pm \sqrt{M^2-4N}}{2} \end{align}

So if $\sqrt{5 + 2\sqrt 6} = \sqrt a + \sqrt b$

$$\{a,b\} = \left\{\dfrac{5 \pm \sqrt{5^2 - 4(6)}}{2}\right\} = \{3,2\}$$

$$x^2-5x+6 = (x-2)(x-3)$$

So $$\sqrt{5 + 2\sqrt 6} = \sqrt 2 + \sqrt 3 $$