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Rationalise the denominator and simplify fully:

$$\dfrac{6}{\sqrt{7} + 2}$$

I got the answer $\dfrac{2 \sqrt{7}}{3}$, but didn't get the mark. Is that not fully simplified?

I did $\displaystyle \frac{6}{\sqrt{7} + 2} \times \frac{\sqrt{7}}{\sqrt{7}} = \frac{6 \sqrt{7}}{(7 + 2)} = \frac{6 \sqrt{7}}{9} = \frac{2 \sqrt{7}}{ 3}$ .

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  • $\begingroup$ Is this $\frac{6}{\sqrt 7 + 2}$ or $\frac{6}{\sqrt 7} + 2$ ? $\endgroup$ – AlexanderJ93 Apr 9 '17 at 1:24
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    $\begingroup$ And if what you meant was $\frac{6}{\sqrt 7 + 2}$ then you either need to use the correct MathJax command, or you need parentheses to group the denominator. $\endgroup$ – quasi Apr 9 '17 at 1:25
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$$\frac{6}{\sqrt{7}+2}=\frac{6(2-\sqrt{7})}{(2+\sqrt{7})(2-\sqrt{7})} = \frac{12-6\sqrt{7}}{4-7} = \frac{3(4-2\sqrt{7})}{-3} = 2\sqrt{7}-4$$

Note that in general $$\frac{a}{b+c\sqrt{d}} = \frac{a(b-c\sqrt{d})}{(b+c\sqrt{d})(b-c\sqrt{d})} = \frac{ab-ac\sqrt{d}}{b^2-c^2d}$$

Notice that the denominator is now rational, i.e. in particular for this case it has no surds in it.

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  • $\begingroup$ I was taught you could just multiply by in this case √7 / √7. Would that not work? $\endgroup$ – Charlie Apr 9 '17 at 1:35
  • $\begingroup$ @Charlie That wouldn't work, you'd still be left with a surd in the denominator, if you would show me the working you used to arrive at your answer, I could tell you where you're going wrong. Also note your answer is incorrect (it doesn't equal the original expression). $\endgroup$ – mrnovice Apr 9 '17 at 1:47
  • $\begingroup$ I did 6 / (√7 + 2) x √7 / √7 = 6 √7 / (7 + 2) = 6 √7 / 9 = 2 √7 / 3 $\endgroup$ – Charlie Apr 9 '17 at 1:50
  • $\begingroup$ $\frac{6}{\sqrt{7}+2}\cdot\frac{\sqrt{7}}{\sqrt{7}}=\frac{6\sqrt{7}}{7+2\sqrt{7}}$-do you see the mistake you made? $\endgroup$ – mrnovice Apr 9 '17 at 1:52
  • $\begingroup$ @Charlie yes that's correct, now if you look at my answer, you can see that the general method to rationalise the denominator $\endgroup$ – mrnovice Apr 9 '17 at 1:57
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I'll assume your expression was $\frac{6}{\sqrt{7}+2}$, since it's the only way you obtain that denominator.

$$\frac{6}{\sqrt{7}+2}\times\frac{\sqrt{7}-2}{\sqrt{7}-2}=\frac{6\sqrt{7}-12}{\sqrt{7}^2-2^2}=\frac{6\sqrt{7}-12}{3}={2\sqrt{7}-4}$$

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  • $\begingroup$ Hi mate, I need it to still be expressed as a fraction. $\endgroup$ – Charlie Apr 9 '17 at 1:36
  • $\begingroup$ From the fraction you gave us, this is the simplest form you can have. Why do you need it as a fraction? $\endgroup$ – AspiringMathematician Apr 9 '17 at 1:50
  • $\begingroup$ Because that's how all these kinds of questions are supposed to be answered on GCSE. $\endgroup$ – Charlie Apr 9 '17 at 1:52

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