Question on: JAMB Mathematics - 2017
If \(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) = m + n β 6,
find the values of m and n respectively
find the values of m and n respectively
A
1, β 2
B
β 2, n = 1
C
\(\frac{-2}{5}\), 1
D
\(\frac{2}{3}\)
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Correct Option: B
\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\)= m + nβ6
\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) x \(\frac{\sqrt{3} - 2 \sqrt{2}}{\sqrt{3} - \sqrt{2}}\)
\(\frac{2 \sqrt{3} (\sqrt{3} - 2 \sqrt{2}) - \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}{\sqrt{3}(\sqrt{3} - 2 \sqrt{2}) + 2 \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}\)
\(\frac{2 \times 3 - 4\sqrt{6} - 6 + 2 \times 2}{3 - 2 \sqrt{6} + 2 \sqrt{6} - 4 \times 2}\)
= \(\frac{6 - 4 \sqrt{6} - \sqrt{6} + 4}{3 - 8}\)
= \(\frac{0 - 4 \sqrt{6} - 6}{5}\)
= \(\frac{10 - 5 \sqrt{6}}{5}\)
= β 2 + β6
β΄ m + n\(\sqrt{6}\) = β 2 + β6
m = β 2, n = 1
\(\frac{2 \sqrt{3} - \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}\) x \(\frac{\sqrt{3} - 2 \sqrt{2}}{\sqrt{3} - \sqrt{2}}\)
\(\frac{2 \sqrt{3} (\sqrt{3} - 2 \sqrt{2}) - \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}{\sqrt{3}(\sqrt{3} - 2 \sqrt{2}) + 2 \sqrt{2}(\sqrt{3} - 2 \sqrt{2})}\)
\(\frac{2 \times 3 - 4\sqrt{6} - 6 + 2 \times 2}{3 - 2 \sqrt{6} + 2 \sqrt{6} - 4 \times 2}\)
= \(\frac{6 - 4 \sqrt{6} - \sqrt{6} + 4}{3 - 8}\)
= \(\frac{0 - 4 \sqrt{6} - 6}{5}\)
= \(\frac{10 - 5 \sqrt{6}}{5}\)
= β 2 + β6
β΄ m + n\(\sqrt{6}\) = β 2 + β6
m = β 2, n = 1
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