3
$\begingroup$

I am stuck in the following puzzle and couldn't find a way to approach this.

$\sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{14 + \sqrt{180}}}}$

Please help.

$\endgroup$
  • 1
    $\begingroup$ Write $\sqrt{180} = a\sqrt{5}$, see that $14+\sqrt{180}$ is a square, continue in the same manner for the result in the next outer square root until you reach the end. $\endgroup$ – Daniel Fischer Oct 21 '14 at 10:11
  • $\begingroup$ It evaluates to $\sqrt{5} + 1$. $\endgroup$ – James Harrison Oct 21 '14 at 10:12
7
$\begingroup$

$$ \sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{14 + \sqrt{180}}}} = \sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{(\sqrt{5}+3)^2}}} = \sqrt{5 + \sqrt{5} + \sqrt{3 + \sqrt{5} + \sqrt{5} + 3}} = \sqrt{5 + \sqrt{5} + \sqrt{6+2\sqrt{5}}} = \sqrt{5 + \sqrt{5} + \sqrt{(\sqrt{5}+1)^2}} = \sqrt{5 + \sqrt{5} + \sqrt{5}+1}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1 $$

$\endgroup$
3
$\begingroup$

Hint: $$\eqalign{ & \sqrt{14 + \sqrt {180}} = 3 + \sqrt 5 \cr & \sqrt {3 + \sqrt 5 + 3 + \sqrt 5 } = \sqrt {6 + 2\sqrt 5 } = 1 + \sqrt 5 \cr & \sqrt {5 + \sqrt 5 + \sqrt {3 + \sqrt 5 + 3 + \sqrt 5 } } = \sqrt {6 + 2\sqrt 5 } = 1 + \sqrt 5 \cr} $$

$\endgroup$

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.