Question : If $\frac{\sqrt{26-7 \sqrt{3}}}{\sqrt{14+5 \sqrt{3}}}=\frac{b+a \sqrt{3}}{11}, b>0$, then what is the value of $\sqrt{(\mathrm{b}-\mathrm{a})}$?
Option 1: 5
Option 2: 25
Option 3: 12
Option 4: 9
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Correct Answer: 5
Solution :
Given: $\frac{\sqrt{26-7 \sqrt{3}}}{\sqrt{14+5 \sqrt{3}}}$
$= \frac{\sqrt{26-7 \sqrt{3}}}{\sqrt{14+5 \sqrt{3}}} \times \frac{\sqrt{14-5 \sqrt{3}}}{\sqrt{14-5 \sqrt{3}}} $
$= \frac{\sqrt{364-130\sqrt{3}+105-98\sqrt3}}{\sqrt{196-75}} $
$= \frac{\sqrt{361-228\sqrt{3}+108}}{\sqrt{196-75}} $
$= \frac{\sqrt{19^2+(6\sqrt3)^2-2\times19\times6\sqrt{3}}}{\sqrt{121}} $
$= \frac{\sqrt{(19-6\sqrt3)^2}}{11} $
$= \frac{(19-6\sqrt3)}{11} $
$\therefore \frac{(19-6\sqrt3)}{11}=\frac{b+a \sqrt{3}}{11} $
$b=19, a=-6$
Thus, $\sqrt{(b-a)}=\sqrt{(19-(-6))}=\sqrt{25}=5$
Hence, the correct answer is 5.
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