0

My exams are approaching fast and this question was in one of the sample papers .

I have to simplify $$\frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} + \frac{3\sqrt{2}}{\sqrt{6 + \sqrt{3}}} - \frac{4\sqrt{3}}{\sqrt{6 + \sqrt{2}}}$$

I am a ninth grader and we have been taught how to rationalize the denominator but I have no idea how I could simplify this one where there is a root of a root . Even my math teacher couldn't do it and said that she'll have to check it out .

Mark Bennet
  • 100,194
MayankJain
  • 521
  • 1
    Hello, welcome to Math.SE. I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. – Lord_Farin Sep 15 '13 at 15:02
  • 1
    Hint: $\sqrt{6+\sqrt{2}}\sqrt{6-\sqrt{2}} = \sqrt{36-2} = \sqrt{34}$ – Prahlad Vaidyanathan Sep 15 '13 at 15:05
  • @Mayank, by any chance, you meant $\sqrt6+\sqrt3$ not $\sqrt{6+\sqrt3}$ – lab bhattacharjee Sep 15 '13 at 15:11
  • @Lord_Farin Thanks for pointing that out , I am new here so don't know all the rules , I'll take care of this from next time . – MayankJain Sep 15 '13 at 15:29
  • @labbhattacharjee No , That is why I had a problem with the question . I could solve it if it was $\sqrt{6} + \sqrt{3}$ . – MayankJain Sep 15 '13 at 15:30

1 Answers1

1

Things look pretty nice, if the problem is

$$\frac{\sqrt6}{\sqrt3+\sqrt2}+\frac{3\sqrt2}{\sqrt6+\sqrt3}-\frac{4\sqrt3}{\sqrt6+\sqrt2}$$

Multiplying each term with the conjugate of the respective numerator we get,

$$\frac{\sqrt6(\sqrt3-\sqrt2)}{3-2}+\frac{3\sqrt2(\sqrt6+\sqrt3)}{6-3}-\frac{4\sqrt3(\sqrt6-\sqrt2)}{6-2}$$ $$(\text{ as } (\sqrt3-\sqrt2)(\sqrt3+\sqrt2)=(\sqrt3)^2-(\sqrt2)^2=3-2\text{ and so on} )$$

$$=\sqrt6(\sqrt3-\sqrt2)+\sqrt2(\sqrt6+\sqrt3)-\sqrt3(\sqrt6-\sqrt2)$$

$$=\sqrt{18}-\sqrt{12}+\sqrt{12}-\sqrt6-\sqrt{18}+\sqrt6=0$$

lab bhattacharjee
  • 274,582