Let α and β be the roots of the equation \(\rm \frac{1}{x+a+b}=\frac{1}{x}+\frac{1}{a}+\frac{1}{b}\); a ≠ 0, b ≠ 0, x ≠ 0.

Which one of the following is a quadratic equation whose roots are α² and β²?

Given:

Roots of the equation \(\rm \frac{1}{x+a+b}=\frac{1}{x}+\frac{1}{a}+\frac{1}{b}\) are α and β

Where, a ≠ 0, b ≠ 0, x ​≠ 0

Concept Used:

If the roots of a quadratic equation are p and q

Quadratic equation is →  x² - (p + q)x + pq = 0

Calculation:

According to the question,

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

⇒ \(\frac{1}{x+a+b}\) - \(\frac{1}{x}\) = \(\frac{1}{a}+\frac{1}{b}\)

⇒ \(\frac{x-(x+a+b)}{x(x+a+b)}\) = \(\frac{a+b}{ab}\)

⇒ \(\frac{-(a+b)}{x(x+a+b)}\) = \(\frac{a+b}{ab}\)

⇒ x(x + a + b) = -ab

⇒ x² + ax + bx + ab = 0

⇒ x(x + a) + b(x + a)  = 0

⇒ (x + a)(x + b) = 0 

⇒ x = -a and x = -b

Thus, roots of the equation are α = -a and β = -b

⇒ α² = a² and β² = b²

Quadratic equation whose roots are α² and β²

⇒ x² - (a² + b²)x + a²b² = 0

∴ The required quadratic equation is x2 - (a2 + b2)x + a2b2 = 0.