0
$\begingroup$

I'm on one of the more difficult practice problems on Excercise 1-6 in "AoPS:Vol. 1".

Problem £4 : Ex. 1-6. The hint details that we should multiply $$\left(\frac{1}{\sqrt 1 \sqrt 2}\right) $$ by it's conjugate which the hint identifies as itself.

I have tried rationalizing the above fraction by multiplying it's conjugate by itself, however, how can we FOIL the two terms when there appears to only be one?

Overall Questions: 1. How the Conjugate Radical be the same as the original expression? If so, why? 2. How can we rationalize a fraction like this which appears to have only one term, when rationalization by conjugate radicals always requires two?

Thanks.

$\endgroup$
0
$\begingroup$

$1/\sqrt{1+\sqrt{2}} = \sqrt{1+\sqrt{2}}/(1+\sqrt{2})$.

Now multiply top and bottom by $1-\sqrt(2)$.

$\sqrt{1+\sqrt{2}}(1-\sqrt(2))/((1-\sqrt(2))((1+\sqrt{2}))$

$=\sqrt{1+\sqrt{2}}(1-\sqrt(2))/(-1)$

$=\sqrt{1+\sqrt{2}}(\sqrt(2)-1)$

$\endgroup$
  • $\begingroup$ Hmmm,... I seem to be having issues from the very first step! How do we go from the first step to the second one? How does the denominator become (1+sqrt(2)) in the second step? Did we multiply by a conjugate, or...? $\endgroup$ – DarkRunner Jul 22 '16 at 0:32
  • $\begingroup$ @RefathBari: No, XSPX multiplied both the numerator and denominator by $\sqrt{1+\sqrt{2}}$ to obtain $\frac {\sqrt{1+\sqrt{2}}}{1+\sqrt{2}}$ $\endgroup$ – Frank Aug 2 '16 at 21:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.