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

We are given an equation involving the expression: \(\frac{x+1}{x}\). Let's make a substitution:

Let: \(\frac{x+1}{x} = a\)

The original equation becomes: 

\[a^2 - 3a + 2 = 0\]

Solving this quadratic: 

\[a = 1 \quad \text{or} \quad a = 2\]

 Analyze Each Value of \(a\)

• For \(a = 2 \Rightarrow \frac{x+1}{x} = 2\)
  Multiply both sides by \(x\): \(x + 1 = 2x \Rightarrow x = 1\)
  This is a valid real solution ✅
• For \(a = 1 \Rightarrow \frac{x+1}{x} = 1\)
  Multiply both sides by \(x\): \(x + 1 = x \Rightarrow 1 = 0\)
  ❌ Contradiction! So this is not valid.

 Final Result

Therefore, the only real solution is: \(x = 1\)
Number of real 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 \(1\).