Question

If x+1/x=4,find the value of x-1/x​

Answer 1

Question :–

• If $\: { \bold{x + \dfrac{1}{x} = 4}}$ , then find the value of $\: { \bold{x - \dfrac{1}{x} }}$ .

ANSWER :–

$\\ \longrightarrow \: { \boxed{ \bold{x - \dfrac{1}{x} = \pm \sqrt{12} }}}$

EXPLANATION :–

GIVEN :–

• Value of $\: { \bold{x + \dfrac{1}{x} }}$ is 4.

TO FIND :–

• $\: { \bold{x - \dfrac{1}{x} = \: ? }}$

SOLUTION :–

$\\ \implies { \bold{x + \dfrac{1}{x} = 4}}$

• Square both side –

$\\ \implies { \bold{(x + \dfrac{1}{x}) ^{2} = {(4)}^{2} }}$

$\\ \implies { \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } + 2(x)( \frac{1}{x} )= 16 }}$

$\\ \implies { \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } + 2 = 16 }}$

$\\ \implies { \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } = 16 - 2 }}$

$\\ \implies { \bold{ {x}^{2} + \dfrac{1}{ {x}^{2} } = 14 \: \: \: \: \: \: - - - - eq.(1) }}$

• Now –

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

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

$\\ \implies \: { \bold{(x - \dfrac{1}{x} ) ^{2} = 14 - 2 \: \: \: [using \: \: eq.(1)]}}$

$\\ \implies \: { \bold{(x - \dfrac{1}{x} ) ^{2} = 12}}$

• Now take square root on both side –

$\\ \implies \: { \boxed{ \bold{x - \dfrac{1}{x} = \pm \sqrt{12} }}}$

Used formula :–

$\\ \: { \bold{ (1) \: \: {(a + b)}^{2} = {a}^{2} + {b}^{2} + 2ab }}$

$\\ \: { \bold{ (2) \: \: {(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab }}$