KnowledgeBoat Logo
|

Mathematics

If x2+1x2=2x^2 + \dfrac{1}{x^2} = 2, the value of x21x2x^2 - \dfrac{1}{x^2} is :

  1. 0

  2. 2

  3. ±2\pm 2

  4. 8

Expansions

11 Likes

Answer

By formula,

(x2+1x2)2(x21x2)2\Rightarrow \Big(x^2 + \dfrac{1}{x^2}\Big)^2 - \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 4.

Substituting values we get :

22(x21x2)2=44(x21x2)2=4(x21x2)2=44(x21x2)2=0x21x2=0.\Rightarrow 2^2 - \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 4 \\[1em] \Rightarrow 4 - \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 4 \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 4 - 4 \\[1em] \Rightarrow \Big(x^2 - \dfrac{1}{x^2}\Big)^2 = 0 \\[1em] \Rightarrow x^2 - \dfrac{1}{x^2} = 0.

Hence, Option 1 is the correct option.

Answered By

7 Likes


Related Questions