Easy surds

Level 2

If the value of 16 + 18 2 3 2 2 \frac {16 + 18\sqrt{2}}{3 - 2\sqrt{2}} can be simplified as a + b c a + b\sqrt{c} , what is the value of a + b + c a + b + c ?


The answer is 208.

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1 solution

Sharky Kesa
Feb 7, 2014

For this question, you may need to know a bit about conjugates. 16 + 18 2 3 2 2 \frac {16 + 18\sqrt{2}}{3 - 2\sqrt{2}} is equivalent to ( 16 + 18 2 ) ( 3 + 2 2 ) ( 3 2 2 ) ( 3 + 2 2 ) \frac {(16 + 18\sqrt{2})(3 + 2\sqrt{2})}{(3 - 2\sqrt{2})(3 + 2\sqrt{2})} . Solving this we get 120 + 86 2 9 8 \frac {120 + 86\sqrt{2}}{9 - 8} or simply 120 + 86 2 120 +86\sqrt{2} . By adding 120, 86 and 2 we get an answer of 208 208 .

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