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's solve the given equation: \(\left(\frac{x+1}{x}\right)^2 - 3\left(\frac{x+1}{x}\right) + 2 = 0\). .

Step 1: Let \( y = \frac{x+1}{x}\). Substituting, we get the quadratic equation:

\(y^2 - 3y + 2 = 0\).

Step 2: Solve the quadratic equation using the quadratic formula, \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -3, c = 2 \).

\(y = \frac{-(-3) \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot 2}}{2 \cdot 1}\)

\(y = \frac{3 \pm \sqrt{9 - 8}}{2}\)

\(y = \frac{3 \pm 1}{2}\)

This gives two solutions: \(y = 2\) and \(y = 1\).

Step 3: Convert back to \( x \) using \( y = \frac{x+1}{x} \):

Case 1: If \( y = 2\), then:

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

\( x+1 = 2x\)

\( x = 1\).

Case 2: If \( y = 1\), then:

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

\( x+1 = x\)

This implies \( 1=0 \), which is not possible, so there is no solution in this case.

Thus, the only distinct real root is \(x = 1\).

Therefore, the number of distinct real roots is 1.

Answer 2

Assume \( x + \frac{1}{x} = a \) 

Then, the given equation becomes: \( a^2 - 3a + 2 = 0 \)

Solving this quadratic equation:

\( a^2 - 3a + 2 = 0 \Rightarrow (a - 1)(a - 2) = 0 \)

So, \( a = 1 \) or \( a = 2 \)

Case 1: If \( a = 1 \), then \( x + \frac{1}{x} = 1 \)

This implies \( x^2 - x + 1 = 0 \), whose discriminant is: \( D = (-1)^2 - 4(1)(1) = 1 - 4 = -3 \)

Since the discriminant is negative, this equation has no real roots.

Case 2: If \( a = 2 \), then \( x + \frac{1}{x} = 2 \)

This implies \( x^2 - 2x + 1 = 0 \Rightarrow (x - 1)^2 = 0 \)

So, \( x = 1 \) is the only real root.

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