Jambmaths question:
if $\frac{2\sqrt{3}-2}{\sqrt{3}+2\sqrt{2}}=m+n\sqrt{6}$. If the value of m and n respectively
Option A:
1, –2
Option B:
–2, 1
Option C:
$-\tfrac{2}{5}$,1
Option D:
$2,\tfrac{3}{5}$
Jamb Maths Solution:
$\begin{align} & \frac{2\sqrt{3}-2}{\sqrt{3}+2\sqrt{2}}=\frac{2\sqrt{3}-2}{\sqrt{3}+2\sqrt{2}}\times \frac{\sqrt{3}-2\sqrt{2}}{\sqrt{3}-2\sqrt{2}}\text{ } \\ & \text{multiplying the denomination and denominator by } \\ & \sqrt{3}-2\sqrt{2}\text{ which the conjugate of }\sqrt{3}+2\sqrt{2} \\ & =\frac{6-4\sqrt{6}-\sqrt{6}+4}{3-8}=\frac{10-5\sqrt{6}}{-5}=-2+\sqrt{6} \\ & m=-2,n=1 \\\end{align}$
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