Question

If \( \alpha \) and \( \beta \) are the roots of the equation \( ax^2 + bx + c = 0 \), then the equation whose roots are \( \alpha + \beta \) and \( \frac{1}{\alpha} + \frac{1}{\beta} \) is

• A. \( acx^2 - (ab + bc)x + b^2 = 0 \)
• B. \( acx^2 + (ab + bc)x + b^2 = 0 \)
• C. \( acx^2 + b(a + c)x + b^2 = 0 \)
• D. \( acx^2 - b(a + c)x - b^2 = 0 \)

Hint:

When forming a new quadratic equation from the roots of a given quadratic equation, first find the sum and product of the original roots using Vieta's formulas. Then, express the new roots in terms of the original roots and find the sum and product of the new roots. Finally, use the formula \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \) to form the new equation.

Answer 1

The Correct Option is C

Given the quadratic equation \( ax^2 + bx + c = 0 \) with roots \( \alpha \) and \( \beta \), by Vieta's formulas: $$ \alpha + \beta = -\frac{b}{a} $$ $$ \alpha \beta = \frac{c}{a} $$ The roots of the new quadratic equation are \( r_1 = \alpha + \beta \) and \( r_2 = \frac{1}{\alpha} + \frac{1}{\beta} \).
We have \( r_1 = -\frac{b}{a} \).
And \( r_2 = \frac{\alpha + \beta}{\alpha \beta} = \frac{-b/a}{c/a} = -\frac{b}{c} \).
The new quadratic equation is given by \( x^2 - (r_1 + r_2)x + r_1 r_2 = 0 \).
The sum of the new roots is: $$ r_1 + r_2 = -\frac{b}{a} + (-\frac{b}{c}) = -\frac{b}{a} - \frac{b}{c} = -b\left(\frac{1}{a} + \frac{1}{c}\right) = -b\left(\frac{c + a}{ac}\right) = -\frac{b(a + c)}{ac} $$ The product of the new roots is: $$ r_1 r_2 = \left(-\frac{b}{a}\right) \left(-\frac{b}{c}\right) = \frac{b^2}{ac} $$ Substituting these into the quadratic equation form: $$ x^2 - \left(-\frac{b(a + c)}{ac}\right)x + \frac{b^2}{ac} = 0 $$ $$ x^2 + \frac{b(a + c)}{ac}x + \frac{b^2}{ac} = 0 $$ Multiplying by \( ac \) to eliminate the denominators: $$ acx^2 + b(a + c)x + b^2 = 0 $$