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.
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.
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 $a^2 + b^2 + 2ab = (a+b)^2$
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.
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 $$\sqrt{5 + 2\sqrt 6} = \sqrt 2 + \sqrt 3 $$
there are four possible answers
No, there is only one single answer in real numbers.
– dxiv
Feb 08 '18 at 08:36