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How do we find $p$ and $q$ so that $$\frac{\sqrt{2}+2}{\sqrt{2}-1}=p+q\sqrt{2}\,?$$

symplectomorphic
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Kristin
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  • Sorry, I should have clarified. The question asks to find the values of p and q. I have the answer key, which states that p = 4 and q = 3, but do not know how to arrive at that answer. – Kristin Apr 30 '14 at 04:03

1 Answers1

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Just multiply numerator and denominator by the "conjugate" $\sqrt{2}+1$, giving $$\frac{\sqrt{2}+2}{\sqrt{2}-1}\cdot\frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{2+\sqrt{2}+2\sqrt{2}+2}{1}=4+3\sqrt{2}$$

symplectomorphic
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  • Excellent. Thanks. – Kristin Apr 30 '14 at 04:09
  • you're welcome. (you can upvote my answer and select it as the answer to your question to show thanks, too.) this trick of multiplying by the conjugate is very useful when you're doing arithmetic with radicals. – symplectomorphic Apr 30 '14 at 04:10