Question 14

If $$P = 7 + 4\surd3$$ and $$PQ = 1$$, then what is the value of $$\frac{1}{P^2} + \frac{1}{Q^2}$$?

Solution

$$P = 7 + 4\surd3$$
PQ=1
Q=1/($$ 7 + 4\surd3$$)
Rationalizing the denominator i.e multiplying both numerator and denominator with $$ 7 - 4\surd3$$
Q=$$ 7 - 4\surd3$$
$$\frac{1}{P^2} + \frac{1}{Q^2}$$

=$$\frac{P^2+Q^2}{P^2Q^2} $$

=$$\frac{(P+Q)^2-2PQ}{P^2Q^2} $$

P+Q=$$7 + 4\surd3+7 - 4\surd3$$
P+Q=14
PQ=1
Therefore =(196-2)/1
=194

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SSC CGL Tier-2-17-February-2018 Maths

If $$P = 7 + 4\surd3$$ and $$PQ = 1$$, then what is the value of $$\frac{1}{P^2} + \frac{1}{Q^2}$$?

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