Question

The number of distinct real roots of the equation \({\left( {x + \frac{1}{x}} \right)^2} - 3\left( {x + \frac{1}{x}} \right) + 2 = 0\) equals

Answer 1

Calculation:

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

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

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

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

⇒ \({\left( {x + \frac{1}{x}} - 1 \right)} = 0\ and \ \left( {x + \frac{1}{x}} - 2 \right) = 0\)

Let's solve  \({\left( {x + \frac{1}{x}} - 1 \right)} = 0\)

⇒ x² + 1 - x = 0

⇒ x² - x +1 = 0

Roots of the equation

⇒ D = b² - 4ac

⇒ D = (-1)² - 4 × 1 × 1

⇒ D = -3

So, roots are Imagenary,

Now lets solve \( \ \left( {x + \frac{1}{x}} - 2 \right) = 0\)

⇒ x² + 1 - 2x = 0

⇒ x² - 2x + 1 = 0

Roots of the equations

⇒ D = b² - 4ac

⇒ D = (-2)² - 4 × 1 × 1

⇒ D = 0

So, roots are Real and Equal

⇒ x² - 2x + 1 = 0

⇒ (x - 1)² = 0

⇒ x = 1

∴ The distinct roos of the equation is 1.