3
$\begingroup$

What is the value of $\cfrac{\sqrt8+\sqrt{27}}{5-\sqrt6}-2(\sqrt[4]9-1)^{-1}$ ? $$1)1+\sqrt3\quad\quad\quad\quad\quad\quad2)-1+\sqrt2\quad\quad\quad\quad\quad\quad3)1-\sqrt2\quad\quad\quad\quad\quad\quad4)\sqrt2-2\sqrt3$$

It was one of the questions from timed exam (for example supposed to solve any question in average one minute). but I have difficulty to calculate this expression fast. my approach is:

$$\left(\frac{2\sqrt2+3\sqrt3}{5-\sqrt6}\times\frac{5+\sqrt6}{5+\sqrt6}\right)-\frac2{\sqrt3-1}$$ By rationalizing the second fraction it is not hard to see it is $\sqrt3+1$. but I had difficulty to accurately calculate the first one in short time. Is it possible to evaluate it quicker?

$\endgroup$
7
  • $\begingroup$ I'm getting $\sqrt 2-1$. Given that it is multiple choice you can just estimate the number...the answers are far apart. WA confirms my answer. $\endgroup$
    – lulu
    Feb 24 '21 at 0:36
  • $\begingroup$ I guess you can memorise 1.41 and 1.73? Not sure how fast this will be though. Try it. Like lulu said, this is motivated by the lack of nice cancellations, and the far distances between answers. $\endgroup$ Feb 24 '21 at 0:38
  • $\begingroup$ It took me a minute and 39 seconds to get $\sqrt2-1$ as the answer. I'm not sure how I could have done it any faster. $\endgroup$ Feb 24 '21 at 0:40
  • $\begingroup$ @BenjaminWang Ok but $\sqrt6$ is a little trickier to memorize and I think a lot of people haven't memorized it. but if someone memorized $24^2=576$ and $25^2=625$ it is not hard to recognize it is estimately $2.45$ ;) $\endgroup$
    – Etemon
    Feb 24 '21 at 0:40
  • $\begingroup$ @BarryCipra It is ok. I just a little exaggerated when I said one minute ( I considered having stress in the exam and the time it takes to fill in the answer sheet and so on) by the way it was the first question! $\endgroup$
    – Etemon
    Feb 24 '21 at 0:55
2
$\begingroup$

If you are very clever, it might occur to you that

$$\sqrt8+\sqrt{27}=(\sqrt2)^3+(\sqrt3)^3=(\sqrt2+\sqrt3)((\sqrt2)^2-\sqrt2\sqrt3+(\sqrt3)^2)=(\sqrt2+\sqrt3)(2-\sqrt6+3)$$

which lets you cancel out the $5-\sqrt6$ from the denominator. But that only occurred to me after the fact.

$\endgroup$
2
$\begingroup$

$$\begin{align} \frac{\sqrt{8}+\sqrt{27}}{5-\sqrt{6}} &= \frac{(\sqrt{8} + \sqrt{27})(5 + \sqrt{6})}{25-6} \\ &= \frac{5 \sqrt{8} + \sqrt{48} + 5\sqrt{27} + \sqrt{81(2)}}{19} \\ &= \frac{10\sqrt{2} + 4\sqrt{3} + 15\sqrt{3} + 9\sqrt{2}}{19} \\ &= \sqrt{2} + \sqrt{3}. \end{align}$$

$\endgroup$

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.