Question

The number of distinct real roots of the equation \(\bigg(\frac{x+1}{x}\bigg)^2-3\bigg(\frac{x+1}{x}\bigg)+2=0\) equals
[This Question was asked as TITA]

• A. 2
• B. 3
• C. 4
• D. 1

Answer 1

The Correct Option is D

Let \(\frac{x+1}{x} = a\)
The given equation becomes, \(a^2-3a+2 = 0\)  \(a= 2\) or \(1\) i.e. \(\frac{x+1}{x}\) = \(2\) or \(\frac{x+1}{x} = 1\)

since \(x\) is real, \(x+\frac{1}{x} ≠1\); 

∴ \(\frac{x+1}{x} = 2\)

\(∴\) The number of solutions = \(1\)

Answer 2

Assume \(x+\frac{1}{x}= a\)

∴ Given equation is now : \(a^2-3a+2=0\)
So, ‘a’ can be either 1 or 2
If \(a=1\),  \(x+\frac{1}{x}=1\) and it has no real roots.

If \(a=2\), \(x+\frac{1}{x}=2\) and it has exactly one real root which is \(x=1\)

Therefore, the total number of distinct real roots of the given equation is 1