Question : If $\frac{\sqrt{38-5 \sqrt{3}}}{\sqrt{26+7 \sqrt{3}}}=\frac{a+b \sqrt{3}}{23}, b>0$, then the value of $(b-a)$ is:
Option 1: 7
Option 2: 18
Option 3: 29
Option 4: 11
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Correct Answer: 29
Solution : $\frac{\sqrt{38-5 \sqrt{3}}}{\sqrt{26+7 \sqrt{3}}}=\frac{a+b \sqrt{3}}{23}, b>0$ $⇒\frac{\sqrt{38-5 \sqrt{3}}}{\sqrt{26+7 \sqrt{3}}}=\frac{a+b \sqrt{3}}{23}$ $⇒\frac{\sqrt{(38-5 \sqrt{3)}({26-7 \sqrt{3}})}}{\sqrt{(26+7 \sqrt{3})({26-7 \sqrt{3}})}}=\frac{a+b \sqrt{3}}{23}$ $⇒\frac{\sqrt{(988+105-396\sqrt3)}}{\sqrt{(529)}}=\frac{a+b \sqrt{3}}{23}$ $⇒\frac{\sqrt{(11^2+(18\sqrt3)^2-2\times11\times18\sqrt3)}}{\sqrt{(23^2)}}=\frac{a+b \sqrt{3}}{23}$ $⇒\frac{\sqrt{(18\sqrt3-11)^2}}{23}=\frac{a+b \sqrt{3}}{23}$ $⇒\frac{(18\sqrt3-11)}{23}=\frac{a+b \sqrt{3}}{23}$ On comparing, $a=-11$ and $b=18$ So, $(b-a)=18+11 = 29$ Hence, the correct answer is 29.
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