After rationalizing the denominator of frac{7 | vedclass

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Question

(C) To rationalize the denominator of $\frac{7}{3 \sqrt{3}-2 \sqrt{2}},$ we multiply the numerator and the denominator by the conjugate of the denomin

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) To rationalize the denominator of $\frac{7}{3 \sqrt{3}-2 \sqrt{2}},$ we multiply the numerator and the denominator by the conjugate of the denominator,which is $(3 \sqrt{3}+2 \sqrt{2}).$$\frac{7}{3 \sqrt{3}-2 \sqrt{2}} 	imes \frac{3 \sqrt{3}+2 \sqrt{2}}{3 \sqrt{3}+2 \sqrt{2}} = \frac{7(3 \sqrt{3}+2 \sqrt{2})}{(3 \sqrt{3})^2 - (2 \sqrt{2})^2}$$= \frac{7(3 \sqrt{3}+2 \sqrt{2})}{(9 	imes 3) - (4 	imes 2)} = \frac{7(3 \sqrt{3}+2 \sqrt{2})}{27 - 8}$$= \frac{7(3 \sqrt{3}+2 \sqrt{2})}{19}$Thus,the denominator is $19.$Therefore,the correct option is $(C)$.

After rationalizing the denominator of $\frac{7}{3 \sqrt{3}-2 \sqrt{2}},$ we get the denominator as

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