If frac{5+2 sqrt{3}}{7+4 sqrt{3}}=a+b sqrt{3} | vedclass

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(C) To solve $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3},$ we rationalize the denominator by multiplying the numerator and denominator by the conj

Vedclass · Accepted Answer

$a=11, b=-6$

Vedclass · Answer

(C) To solve $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3},$ we rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator,which is $(7-4 \sqrt{3})$.$\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}} 	imes \frac{7-4 \sqrt{3}}{7-4 \sqrt{3}} = \frac{35 - 20 \sqrt{3} + 14 \sqrt{3} - 8(3)}{49 - 16(3)}$$= \frac{35 - 24 - 6 \sqrt{3}}{49 - 48}$$= \frac{11 - 6 \sqrt{3}}{1} = 11 - 6 \sqrt{3}$Comparing $a+b \sqrt{3} = 11 - 6 \sqrt{3}$,we get $a = 11$ and $b = -6$.

If $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3},$ then

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