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# Algebra

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Rationalize (sqrt(2) + sqrt(3))/(sqrt(10) + sqrt(14) + sqrt(15) + sqrt(21))

Feb 14, 2022

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$$\frac{\sqrt2+\sqrt3}{\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}}$$

Looking at the denominator, notice that $$\sqrt{10}$$ and $$\sqrt{14}$$ are both divisible by $$\sqrt{2}$$ and$$\sqrt{15}$$, $$\sqrt{21}$$ are divisible by $$\sqrt{2}$$

We can factor those out to get $$\frac{\sqrt2+\sqrt3}{\sqrt2(\sqrt5+\sqrt7)+\sqrt{3}(\sqrt5+\sqrt7)} = \frac{\sqrt2+\sqrt3}{(\sqrt2+\sqrt{3})(\sqrt5+\sqrt7)}$$. Divide top and bottom by $$\sqrt2+\sqrt3$$$$\frac{1}{\sqrt5+\sqrt7}$$.

Moving the radicals to the numerator: $$\frac{1}{\sqrt5+\sqrt7} = \frac{1\cdot(\sqrt5-\sqrt7)}{(\sqrt5+\sqrt7)(\sqrt5-\sqrt7)} = \frac{\sqrt{7}-\sqrt{5}}{2}$$.

So the final result is: $$\boxed{\frac{\sqrt7 - \sqrt5}{2}}$$. Hope this helped

Feb 14, 2022
edited by asiandude  Feb 14, 2022
edited by asiandude  Feb 14, 2022