Find the value of b :frac{sqrt{2}+sqrt{3}}{3 | vedclass

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(C) Given expression: $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}} = 2 - b\sqrt{6}$Rationalize the denominator by multiplying the numerator and de

Vedclass · Accepted Answer

$-\frac{5}{6}$

Vedclass · Answer

(C) Given expression: $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}} = 2 - b\sqrt{6}$Rationalize the denominator by multiplying the numerator and denominator by $(3\sqrt{2} + 2\sqrt{3})$:$= \frac{(\sqrt{2}+\sqrt{3})(3\sqrt{2}+2\sqrt{3})}{(3\sqrt{2}-2\sqrt{3})(3\sqrt{2}+2\sqrt{3})}$$= \frac{3(\sqrt{2})^2 + 2\sqrt{6} + 3\sqrt{6} + 2(\sqrt{3})^2}{(3\sqrt{2})^2 - (2\sqrt{3})^2}$$= \frac{3(2) + 5\sqrt{6} + 2(3)}{18 - 12}$$= \frac{6 + 5\sqrt{6} + 6}{6} = \frac{12 + 5\sqrt{6}}{6} = 2 + \frac{5}{6}\sqrt{6}$Comparing $2 + \frac{5}{6}\sqrt{6}$ with $2 - b\sqrt{6}$:$-b = \frac{5}{6} \Rightarrow b = -\frac{5}{6}$

Find the value of $b$ :
$\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}$

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Find the value of $b$ :$\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}$