Question

If X minus one upon X equals two minus one then what is the value of X square plus one upon X Square?

Answer 1

Given

$x\:-\: \dfrac{1}{x} \:=\: 2\:-\:1$ ......eqn(i)

To Find

The value of $x^2\:+\: \dfrac{1}{x^2}$

Formulas

• (a + b)² = a² + b² + 2ab

• (a - b)² = a² + b² - 2ab

• a² - b² = (a + b) (a - b)

• a² + b² = ( a + b)² - 2ab or ( a - b)² + 2ab

Calculations

Simplifying the RHS in eqn(i)

$x\:-\: \dfrac{1}{x} \:=\: 1$

Squaring both sides, we get :-

$\implies \:{(x\:-\: \dfrac{1}{x}) }^2 \:=\: {1}^2 \\ \\ \\ \implies \: x^2 \:+\: \dfrac{1}{x^2}\:-\:2\:=\:1 \\ \\ \\ \implies x^2 \:+\: \dfrac{1}{x^2} \:=\: 1\:+\:2 \\ \\ \\ \implies x^2\:+\: \dfrac{1}{x^2}\:=\:3$

Hence,the required answer is 3.

Answer 2

Answer:

$x - \frac{1}{x} = 2 - 1 = 1 \\ \\ = \\ \\ ({x - \frac{1}{x} })^{2} = {1}^{2} = 1 \\ \\ = > 1 = {x}^{2} + \frac{1}{ {x}^{2} } - 2 \\ \\ = > {x}^{2} + \frac{1}{ {x}^{2} } = 1 + 2 = 3 \: \:( answer)$