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The question I was working on was to rationalize the denominator of $$\frac{1}{\sqrt{3}(\sqrt{21}+\sqrt{7})}$$

My answer was $\frac{\sqrt7}{42}(3-\sqrt{3})$.

But both my book and Symbolab gave the answer as $\frac{1}{42}(3\sqrt{7}-\sqrt{21})$.

Why shouldn't I factor out the $\sqrt{7}$?

Thanks for your help.

Chinnapparaj R
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ALLearner
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    https://math.stackexchange.com/questions/2047349/when-does-sqrta-b-sqrta-sqrtb https://math.stackexchange.com/questions/1274936/proving-sqrtab-sqrt-a-sqrt-b – lab bhattacharjee Dec 05 '18 at 12:55
  • I think that, since you need to rationalize the denominator, both your solution and Symbolab answer are perfectly fine. – rafa11111 Dec 05 '18 at 13:00

1 Answers1

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The book just wrote the answer in another form.

$$\frac{\sqrt{7}}{42}\cdot(3-\sqrt{3}) = \frac{1}{42}\cdot\sqrt 7\cdot(3-\sqrt 3) = \frac{1}{42}\cdot(3\sqrt 7-\sqrt{21})$$

I think your answer is fine as well. You’ve rationalized the denominator, which is precisely what the question asked for. Whether you want to leave your answer as it is or “play around” with it is a matter of choice.

As another note, I think you started off by factoring $\sqrt 7$ in the denominator while the solutions given probably involved rationalizing immediately. It makes no difference anyway and both are acceptable.

KM101
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