Question

Rationalize the denominator of $\displaystyle \frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}$, and write your answer in the form\[

Answer 1

9514 1404 393

Answer:

  57

Step-by-step explanation:

Apparently, you want to simplify ...

  $\displaystyle \frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}$

so the denominator is rational. It looks like the form you want is ...

  $\dfrac{A\sqrt{2} + B\sqrt{3} + C\sqrt{7} + D\sqrt{E}}{F}$

And you want to know the sum A+B+C+D+E+F.

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We can start by multiplying numerator and denominator by a conjugate of the denominator. Then we can multiply numerator and denominator by a conjugate of the resulting denominator.

  $\displaystyle =\frac{1}{\sqrt{2} + \sqrt{3} + \sqrt{7}}\cdot\frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{\sqrt{2} + \sqrt{3} - \sqrt{7}}=\frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{2\sqrt{6}-2}\\\\=\frac{\sqrt{2} + \sqrt{3} - \sqrt{7}}{2\sqrt{6}-2}\cdot\frac{\sqrt{6}+1}{\sqrt{6}+1}=\frac{(1+\sqrt{6})(\sqrt{2}+\sqrt{3}-\sqrt{7})}{10}\\\\=\frac{\sqrt{2}+\sqrt{3}-\sqrt{7}+2\sqrt{3}+3\sqrt{2}-\sqrt{42}}{10}=\frac{4\sqrt{2}+3\sqrt{3}-\sqrt{7}-\sqrt{42}}{10}$

Comparing this to the desired form we have ...

  A = 4, B = 3, C = -1, D = -1, E = 42, F = 10

Then the sum is ...

  A +B +C +D +E +F = 4 + 3 -1 -1 +42 +10 = 59 -2 = 57

The sum of interest is 57.