Solution
squaring both sides,
$$x+\frac{1}{x}-2=18$$ $$[(a-b)^{2}=a^{2}+b^{2}-2ab]$$
$$\Rightarrow$$ $$x+\frac{1}{x}=20$$
squaring again both sides,
$$\Rightarrow$$ $$x^{2}+\frac{1}{x^{2}}+2=20^{2}$$
$$\Rightarrow$$ $$x^{2}+\frac{1}{x^{2}}=400-2$$ = 398
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
