I have an answer to a problem that I am working on but I have no idea how to rationalize the denominator because I have never worked with this type of problem before. The problem is:
$\frac{\sqrt{2-\sqrt{2}}}{\sqrt{2+\sqrt{2}}}$.
I know that I am supposed to multiply the entire fraction by the denominator but I do not know how it distributes over. How would I go about simplifying these types of problems?
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1 Answer
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Note that you could just rewrite this as: $$\sqrt{\frac{2 - \sqrt{2}}{2 + \sqrt{2}}}$$
Now use your regular techniques for rationalizing what's under the radical sign, get an answer, and put it back under the radical sign!
I should note, though, that the final expression you end up with in this way can still be simplified further, but that requires a bit of trickery.
In particular, you end up with $\sqrt{3 - 2 \sqrt{2}}$. Can this be simplified further? In other words, is there anything you can square to get $3 - 2\sqrt{2}$?
If so, you might imagine it'd be something like $(a + b\sqrt{2})^2$.
But this expression is equal to $a^2 + 2b^2 + 2ab\sqrt{2}$.
We now want: $a^2 + 2b^2 = 3$ and $2ab = -2$.
With two variables and two equations, we can solve for $a$ and $b$, and we find:
$a = 1, b = -1$; or $a = -1, b = 1$.
Thus, the two square roots are: $1 - \sqrt{2}$ and $-1 + \sqrt{2}$.
answered Nov 5, 2012 at 21:30
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$\begingroup$ How would I multiply the radicals together if the signs are different? Am I going to use foil? ie: [sqrt(2)-sqrt(2)]*[sqrt(2)+sqrt(2)] $\endgroup$– KotCommented Nov 5, 2012 at 21:37
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$\begingroup$ The key idea behind rationalizing is that you can multiply by $1$ without changing a number's value. In this particular case, the expression under the radical is rationalized by multiplying by $1 = (2 - \sqrt{2})/(2 - \sqrt{2})$. $\endgroup$ Commented Nov 5, 2012 at 21:40
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$\begingroup$ I realized that but what happens in the denominator? Do I have to use FOIL to cancel the terms out? $\endgroup$– KotCommented Nov 5, 2012 at 21:42
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$\begingroup$ the denominator becomes $(2 + \sqrt{2})(2 - \sqrt{2})$. Yeah, use FOIL and see what you get! $\endgroup$ Commented Nov 5, 2012 at 21:49