6
$\begingroup$

I need to simplify $\sqrt{9+\sqrt{5}}$

I already do this (proven it) $\sqrt{9-4\sqrt{5}}=2- \sqrt{5}$

But I couldn't when apply to $\sqrt{9+\sqrt{5}}=\sqrt{9-4\sqrt{5}+5\sqrt{5}}=\sqrt{(2+\sqrt{5})^2+5\sqrt{5}}$

PLEASE help me out

$\endgroup$
5
  • $\begingroup$ Note that $\sqrt{x^2+y^2} \neq x+y$. $\endgroup$ – Emily Dec 3 '13 at 21:28
  • 7
    $\begingroup$ The expression is as simple as it gets. Why would you expect it simplifies more? $\endgroup$ – Andrés E. Caicedo Dec 3 '13 at 21:29
  • $\begingroup$ so that i prove that $\sqrt{9+\sqrt{5}}*(2-\sqrt{5})=1$ $\endgroup$ – user113224 Dec 3 '13 at 21:32
  • $\begingroup$ This is not true. The actual identity is $\sqrt{9+4\sqrt5}\cdot(2-\sqrt5)=1$. $\endgroup$ – Andrés E. Caicedo Dec 3 '13 at 21:34
  • $\begingroup$ okey thanks andres i know how to solve it $\endgroup$ – user113224 Dec 3 '13 at 21:38
4
$\begingroup$

Suppose $\sqrt{9+\sqrt{5}} = a+\sqrt{b} $.

Squaring both sides, $9+\sqrt{5} = a^2+b+2a\sqrt{b} = a^2+b+\sqrt{4a^2b} $.

Equating the parts, $9 = a^2+b$ and $5 = 4a^2b$.

From the second, $a^2 = 5/(4b)$, so, from the first, $9 = 5/(4b)+b$, or $4b^2-36b+5 = 0$.

The discriminant is $d = 36^2-4\cdot 4\cdot 5 =16(9^2-5) =16\cdot 76 =64\cdot 39 $. This is not a square of an integer, so there is no integer (or rational) expression in a simplified form.

You could, of course, write $\sqrt{9+\sqrt{5}} = 3\sqrt{1+\sqrt{5}/9} $, but this doesn't seem to be worth much.

$\endgroup$
3
$\begingroup$

Let $\sqrt{9+\sqrt{5}}=A+B \sqrt{5}$. Square each side: $9+\sqrt{5} = A^2 + 2AB \sqrt{5} + 5 B^2$. Now we get two equations and two unknowns and solve for $A$ and $B$...

$\endgroup$
3
  • $\begingroup$ i don't know how to solve for a and b $\endgroup$ – user113224 Dec 3 '13 at 21:31
  • $\begingroup$ @user113224: $$A^2+5B^2=9\;\;,\;\;2AB=1$$ $\endgroup$ – DonAntonio Dec 3 '13 at 21:32
  • $\begingroup$ In general, this is how you approach the problem. In this case, it turns out that there are no integer or rational solutions, so your original equation is as good as it gets. $\endgroup$ – abnry Dec 3 '13 at 21:32

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.