express as rational denominator:

4 / root2 + root3 + root7

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Question

express as rational denominator:
4 / root2 + root3 + root7

Ankush Jain · Accepted Answer

$\frac{4}{\sqrt{2} + \sqrt{3} + \sqrt{7}} = \frac{4}{(\sqrt{2} + \sqrt{3}) + \sqrt{7}}$

Rationalize the denominator, we get

$\frac{4}{(\sqrt{2} + \sqrt{3}) + \sqrt{7}} \times \frac{(\sqrt{2} + \sqrt{3}) - \sqrt{7}}{(\sqrt{2} + \sqrt{3}) - \sqrt{7}} = \frac{4 (\sqrt{2} + \sqrt{3} - \sqrt{7})}{(\sqrt{2} + \sqrt{3})^{2} - 7} = \frac{4 (\sqrt{2} + \sqrt{3} - \sqrt{7})}{2 + 3 + 2 \sqrt{6} - 7} = \frac{4 (\sqrt{2} + \sqrt{3} - \sqrt{7})}{2 \sqrt{6} - 2} = \frac{2 (\sqrt{2} + \sqrt{3} - \sqrt{7})}{\sqrt{6} - 1}$

Again, rationalize the denominator, we get

$= \frac{2 (\sqrt{2} + \sqrt{3} - \sqrt{7})}{\sqrt{6} - 1} \times \frac{\sqrt{6} + 1}{\sqrt{6} + 1} = \frac{2 (\sqrt{2} + \sqrt{3} - \sqrt{7}) (\sqrt{6} + 1)}{6 - 1} = \frac{2}{5} (\sqrt{6} + 1) (\sqrt{2} + \sqrt{3} - \sqrt{7})$

express as rational denominator: 4 / root2 + root3 + root7

express as rational denominator:

4 / root2 + root3 + root7