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
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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