How would I rationalize the following Fraction?
$$ \frac {2}{5-\sqrt2+\sqrt3}$$
I have considered the idea of multiplying by the same radicals, but the 5 prevents that.
Asked
Active
Viewed 318 times
3 Answers
2
Multiply top and bottom by all the "relatives" $5+\sqrt{2}-\sqrt{3}$, $5+\sqrt{2}+\sqrt{3}$ and $5-\sqrt{2}-\sqrt{3}$.
The new denominator is invariant under replacement of $\sqrt{2}$ by $-\sqrt{2}$, also under replacement of $\sqrt{3}$ by $-\sqrt{3}$, so it must be rational.
André Nicolas
- 507,029
-
+1 - the 'baby Galois' view is a really nice way of looking at this. – Steven Stadnicki Mar 15 '14 at 05:24
0
Hint $\ $ Rationalize the denominator using the product of the terms below
$\quad \begin{eqnarray}(5\!+\!\sqrt3-\sqrt2)(5\!+\!\sqrt3+\sqrt2) &\,=\,& (5\!+\!\sqrt3)^2-2 &\,=\,&26+10\sqrt3\\ (5\!-\!\sqrt3-\sqrt2)(5\!-\!\sqrt3+\sqrt2) &=& (5\!-\!\sqrt3)^2-2 &=&26-10\sqrt3\end{eqnarray}\Bigg\rbrace$ multiplied $ \,=\, \ldots$
Bill Dubuque
- 272,048
0
$$\begin{align}\frac{2}{5-\sqrt2+\sqrt3}\left(\frac{5+\sqrt2-\sqrt3}{5+\sqrt2-\sqrt3}\right)&=\frac{10+2\sqrt2-2\sqrt3}{20+2\sqrt6}\\ &=\frac{10+2\sqrt2-2\sqrt3}{20+2\sqrt6}\left(\frac{20-2\sqrt6}{20-2\sqrt6}\right)\\ &=\frac{200-20\sqrt6+40\sqrt2-4\sqrt{12}-40\sqrt3+4\sqrt{18}}{400-24}\\ &=\frac{200-20\sqrt6+52\sqrt2-48\sqrt3}{376}\\ &=\frac{50-5\sqrt6+13\sqrt2-12\sqrt3}{94}\\ &\approx.37609 \end{align}$$ You can check you answer here
homegrown
- 3,678
-
Thanks! I'm writing a program to rationalize radicals, I think this is how I'll do it :) – BillyK Mar 17 '14 at 13:45
-
@BillyK No problem. On MSE, if one answer is the most helpful, we accept it. – homegrown Mar 17 '14 at 15:19