Let \alpha and \beta be the roots of the equation \frac{1}{x+a+b}=\frac{1}{x}+\frac{1}{a}+\frac{1}{b}; a \neq 0, b \neq 0, x \neq 0. Which one of the following is a quadratic equation whose roots are \alpha^{2} and \beta^{2}?
- A. x^{2}+(a^{2}+b^{2})x+a^{2}b^{2}=0
- B. x^{2}-(a^{2}+b^{2})x+a^{2}b^{2}=0 ✓
- C. x^{2}-(a^{2}+b^{2})x-a^{2}b^{2}=0
- D. x^{2}+(a^{2}+b^{2})x-a^{2}b^{2}=0
Correct Answer: B. x^{2}-(a^{2}+b^{2})x+a^{2}b^{2}=0
Explanation
From the given equation, \frac{1}{x+a+b} - \frac{1}{x} = \frac{1}{a} + \frac{1}{b} \implies \frac{x-(x+a+b)}{x(x+a+b)} = \frac{a+b}{ab} \implies \frac{-(a+b)}{x^2+(a+b)x} = \frac{a+b}{ab}. This gives x^2+(a+b)x = -ab \implies x^2+(a+b)x+ab = 0. The roots of this equation are -a and -b. Thus, \alpha = -a and \beta = -b. The new roots are \alpha^2 = a^2 and \beta^2 = b^2. The required quadratic equation is x^2 - (a^2+b^2)x + a^2b^2 = 0.
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