-1
$\begingroup$

I have a simplification question. It may be simple, but I can't solve it. Can someone give an idea ? $$\sqrt{4+\sqrt5}-\sqrt{\frac{4+\sqrt{11}}{2}}$$ I tried to rewrite it $$\sqrt{4+\frac22\sqrt5}-\sqrt{\frac{4+\sqrt{11}}{2}}$$ but can't go further. I am not looking for a complete solution, but just a clue.

$\endgroup$
  • $\begingroup$ Already looks quite simple to me... $\endgroup$ – Lord Shark the Unknown Mar 16 '18 at 11:22
  • $\begingroup$ @LordSharktheUnknown : Can you give me a little hint? $\endgroup$ – Khosrotash Mar 16 '18 at 11:23
  • 1
    $\begingroup$ Try to solve $4+\sqrt{5}=(a+b\sqrt{5})^2$ for $a$ and $b$ and you'll see ... $\endgroup$ – Qurultay Mar 16 '18 at 11:27
  • $\begingroup$ Why do you think simplification is possible? $\endgroup$ – Gerry Myerson Mar 16 '18 at 11:38
  • 2
    $\begingroup$ @Gerry Myerson: since Wolfram Alpha answer is $\sqrt{\dfrac{4-\sqrt{11}}{2}}$ ... unexpected result as for my intuition... $\endgroup$ – Oleg567 Mar 16 '18 at 11:42
2
$\begingroup$

$1$st intuitive step is to see how squared expression will look like (to kill part of square roots): $$ \left(\sqrt{4+\sqrt{5}} - \sqrt{\dfrac{4+\sqrt{11}}{2}}\right)^2 \\ = 4+\sqrt{5} + 2 + \dfrac{\sqrt{11}}{2} - \sqrt{32+8\sqrt{5}+8\sqrt{11}+2\sqrt{55}}; $$

$2$nd intuitive step is try to find full square under the square root: $$ 32+8\sqrt{5}+8\sqrt{11}+2\sqrt{55}=\left(a+b\sqrt{5}+c\sqrt{11}\right)^2 $$

Further steps are for you (small hint: $a,b,c$ are positive integer numbers).

$\endgroup$
2
$\begingroup$

Try to find $\sqrt{4+\sqrt{5}}$: Suppose $(a+b\sqrt{5})^2 = \sqrt{4+\sqrt{5}}.$ Then

$$a^2+5b^2 +2ab\sqrt{5} = 4 +\sqrt{5}.$$

So $2ab = 1$ and $a^2+5b^2 = 4$. Then $b=\frac{1}{2a}$ and we have

$$a^2 + 5\left(\frac{1}{2a}\right)^2 = 4$$

or

$$a^4 -4a^2+\frac{5}{4}$$

which gives us $$a^2 = \frac{4\pm \sqrt{11}}{2}.$$

I choose the '+'. Then

$$b^2 = \frac{1}{4a^2} = \frac{1}{2(4+\sqrt{11})} = \frac{4-\sqrt{11}}{10}.$$

So then

$$\sqrt{4+\sqrt{5}} - \sqrt{\frac{4+\sqrt{11}}{2}} = a+b\sqrt{5} - a = \sqrt{\frac{4-\sqrt{11}}{10}}\sqrt{5} = \sqrt{\frac{4-\sqrt{11}}{2}} .$$

$\endgroup$
1
$\begingroup$

Let $\sqrt{4+\sqrt{5}}=\sqrt{(a+b)^2}=\sqrt{a^2+b^2+2ab}$ with $a^2+b^2=4$ and $2ab=\sqrt{5}$.

$${\begin{cases}a^2+b^2=4\\2ab=\sqrt{5}\end{cases}\Rightarrow {\begin{cases}a^2+b^2=4\\4a^2b^2=5\end{cases}}\Rightarrow {\begin{cases}a^2+b^2=4\\a^2b^2=1.25\end{cases}}}\Rightarrow a^2(4-a^2)=1.25$$

Choose just one correct solution for $a$ and $b$, you can now symbolize the expression.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.