Please enter the following details
Free Daily Targets & Mocks
Study Material & Formulas PDFs
1 on 1 Mentorship
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/($$Â 7 + 4\surd3$$)
Rationalizing the denominator i.e multiplying both numerator and denominator with $$ 7 - 4\surd3$$
Q=$$Â 7 - 4\surd3$$
$$\frac{1}{P^2} + \frac{1}{Q^2}$$
=$$\frac{P^2+Q^2}{P^2Q^2} $$
=$$\frac{(P+Q)^2-2PQ}{P^2Q^2} $$
P+Q=$$7 + 4\surd3+7 - 4\surd3$$
P+Q=14
PQ=1
Therefore =(196-2)/1
=194
Create a FREE account and get:
- Free SSC Study Material - 18000 Questions
- 230+ SSC previous papers with solutions PDF
- 100+ SSC Online Tests for Free
