Question

Q.33) If (x+ 1/x) : (x - 1/x) = 5:3, then the
value(s) of x is/are:
यदि (x+1/x): (x- 1/x) = 5:3 तोx का मान हैं-
[a] +1
[b] + 2
[c] +3
[d] 0
R
​

Answer 1

GIVEN :-

$\\ \sf \: \frac{ x + \dfrac{1}{x} }{x - \dfrac{1}{x} } = \frac{5}{3}$

TO FIND :-

• Value of x.

SOLUTION :-

$\\ \implies \sf \: \dfrac{ \left( x + \dfrac{1}{x} \right)}{\left(x - \dfrac{1}{x} \right)} = \dfrac{5}{3} \\ \\ \\ \implies \sf \: 3\left(x + \dfrac{1}{x} \right) = 5\left( x - \dfrac{1}{x} \right) \\ \\ \\ \implies \sf \: 3x + \dfrac{3}{x} = 5x - \dfrac{5}{x}$

$\implies \sf \: 3x - 5x = - \dfrac{5}{x} - \dfrac{3}{x} \\ \\ \\ \implies \sf \: - 2x = \dfrac{ - 8}{x} \\ \\ \\ \implies\sf \: - 2 {x}^{2} = - 8 \\ \\ \\ \implies \sf \: {x}^{2} = \cancel\dfrac{ - 8}{ - 2} \\ \\ \\ \implies \sf \: {x}^{2} = 4 \\ \\ \\ \implies \boxed{ \sf \: x = \pm \: 2}$

VERIFICATION :-

For x=2 ,

$\\ \mapsto \sf \: \dfrac{2 + \dfrac{1}{2} }{2 - \dfrac{1}{2} } = \dfrac{5}{3} \\ \\ \\ \mapsto\sf \: \dfrac{ \dfrac{4 + 1}{2} }{ \dfrac{4 - 1}{2} } = \dfrac{5}{3}$

$\mapsto\sf \: \dfrac{ \dfrac{5}{ \cancel2} }{ \dfrac{3}{ \cancel2} } = \dfrac{5}{3} \\ \\ \\ \mapsto\sf \: \dfrac{5}{3} = \dfrac{5}{3} \: \: \: \: \: (verified)$

For x = -2 ,

$\\ \mapsto \sf \: \dfrac{ - 2 +\left( - \dfrac{1}{2} \right) }{ - 2 - \left( - \dfrac{1}{2} \right)} = \dfrac{5}{3} \\ \\ \\ \mapsto \sf \: \dfrac{ - 2 - \dfrac{1}{2} }{ - 2 + \dfrac{1}{2} } = \dfrac{5}{3} \\ \\ \\ \mapsto\sf \: \dfrac{ \dfrac{ - 4 - 1}{2} }{ \dfrac{ - 4 + 1}{2} } = \dfrac{5}{3}$

$\mapsto\sf \: \dfrac{ \dfrac{ - 5}{ \cancel2} }{ \dfrac{ - 3}{ \cancel2} } = \dfrac{5}{3} \\ \\ \\ \mapsto\sf \: \dfrac{5}{3} = \dfrac{5}{3} \: \: \: \: (verified)$