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Question 12
If $$P = 7 + 4\surd3$$ and $$PQ = 1$$, then what is the value of $$\left( \frac{1}{P^2} \right) + \left(\frac{1}{Q^2}\right)$$?
Solution
$$\left( \frac{1}{P^2} \right) + \left(\frac{1}{Q^2}\right)$$=$$(\frac{P^{2}+Q^{2}}{P^{2}Q^{2}})$$
=$$(\frac{(P+Q)^{2}-2PQ}{P^{2}Q^{2}})$$
$$P = 7 + 4\surd3$$Â
$$Q = 1/P$$
$$Q=7 - 4\surd3$$
P+Q=14
$$(P+Q)^{2}$$=196
PQ=1
$$(\frac{(P+Q)^{2}-2PQ}{P^{2}Q^{2}})$$=196-2
=194
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