Question : If $P=7+4\sqrt3$ and $PQ=1$, then what is the value of $\frac{1}{P^{2}}+\frac{1}{Q^{2}}$?
Option 1: 196
Option 2: 194
Option 3: 206
Option 4: 182
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Correct Answer: 194
Solution : Given: $PQ=1, P=7+4\sqrt3$ ⇒ $Q=\frac{1}{P}$ ⇒ $P^2=(7+4\sqrt3)^2=97+56\sqrt3$ Taking the inverse on both sides and rationalising the denominators, ⇒ $\frac{1}{P^2}={97 -56\sqrt{3}}$ ⇒ $\frac{1}{P^2}+ \frac{1}{Q^2}= \frac{1}{P^2}+ {P^2}$ ⇒ $\frac{1}{P^2}+ \frac{1}{Q^2}=(97 -56\sqrt{3} +97+56\sqrt{3})$ ⇒ $\frac{1}{P^2}+ \frac{1}{Q^2}= 194$ Hence, the correct answer is 194.
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