5
$\begingroup$

I would like to understand how to remove the root squares from this expression:

$$x = \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}$$

How to do it?

$\endgroup$
1
  • $\begingroup$ The right expression is the actual one. Thanks. $\endgroup$
    – Zistoloen
    Commented Dec 28, 2014 at 13:46

4 Answers 4

7
$\begingroup$

Its all about rationalization,

\begin{align} \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}} &= \frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}} \cdot \frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} - \sqrt{5}} \\[10pt] &=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2} \\[10pt] &=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{(2+3+2\sqrt{6})-(5)} \\[10pt] &=\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{2\sqrt{6}}\cdot\frac{\sqrt6}{\sqrt6} \\[10pt]&=\frac{\sqrt{12}+ \sqrt{18} - \sqrt{30}}{12} \\[10pt]&=\frac{2\sqrt{3} + 3\sqrt{2} - \sqrt{30}}{12} \end{align}

$\endgroup$
2
  • 3
    $\begingroup$ Thanks for the entire calculus. $\endgroup$
    – Zistoloen
    Commented Dec 28, 2014 at 14:14
  • $\begingroup$ In the 4th line, you could have just multiplied by $\frac{\sqrt{6}}{\sqrt{6}}$ instead $\endgroup$
    – Dylan
    Commented Dec 29, 2014 at 3:47
3
$\begingroup$

Here's a push

You can try rationalising it by multiplying the expression by its conjugate.

$$\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}\times\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} - \sqrt{5}}=\frac {\sqrt{2}+ \sqrt{3}- \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}$$ and you can continue it.

$\endgroup$
0
$\begingroup$

$\frac 1{\sqrt{2} + \sqrt{3} + \sqrt{5}}=\frac {\sqrt{2}+ \sqrt{3}- \sqrt{5}}{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}=\frac{\sqrt{2}+ \sqrt{3} - \sqrt5}{2\sqrt{6}}$ I think you can do the rest.

$\endgroup$
2
  • $\begingroup$ Can you explain a little more how you get the second expression from the first one? $\endgroup$
    – Zistoloen
    Commented Dec 28, 2014 at 13:47
  • $\begingroup$ I multiplied numerator and denominator with $\sqrt{2}+\sqrt{3}-\sqrt{5}$. It is well-known trick, that if you have $\sqrt{K} \pm \sqrt{L}$ on denominator it is reasonable to multiply up and down of fraction with $\sqrt{K} \mp \sqrt{L}$. Actually you can do the same if you have just $K$ instead of $\sqrt{K}$. $\endgroup$
    – Mihail
    Commented Dec 28, 2014 at 13:52
0
$\begingroup$

Given expression $$\frac{1}{\sqrt 2+\sqrt 3+\sqrt 5}=\frac{\sqrt 5+\sqrt 3-\sqrt 2}{(\sqrt 5+\sqrt 3)^2-2}\\=\frac{\sqrt 5+\sqrt 3-\sqrt 2}{6+2\sqrt{15}}=\frac{(\sqrt 5+\sqrt 3-\sqrt 2)(6-2\sqrt{15})}{-24}$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .