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

Answer 1

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

Given equation:

\(\left( x + \frac{1}{x} \right)^2 - 3 \left( x + \frac{1}{x} \right) + 2 = 0\)

Let \(t = \frac{x + 1}{x}\).

So, the equation becomes:

\(t^2 - 3t + 2 = 0\)

\((t - 1)(t - 2) = 0\)

So, either \(t = 1\) or \(t = 2\).

Now, substituting back \( t \) in terms of \( x \):

1. When \(t = 1\):

\(\frac{x + 1}{x} = 1\)

\(x + 1 = x\)

\(1 = 0\)

This is not possible.

2. When \(t = 2\):

\(\frac{x + 1}{x} = 2\)

\(x + 1 = 2x\)

\(x = 1\)

Therefore, the only solution is \(x = 1\).

Hence, the number of distinct real roots of the equation is \(1\).