Question

If one root of the equation \( ax^2 + bx + c = 0 \) is twice another root, then the equation \( ax^2 + bx + c = 0 \) is:

• A. \( 36b^3 = 343ac^2 \)
• B. \( 36b^3 + 343ac^2 = 0 \)
• C. \( 36b^3 + 729ac^2 = 0 \)
• D. \( 36b^3 = 729ac^2 \)

Hint:

For quadratics with relationships between roots, use Vieta's formulas and manipulate the roots algebraically to find the required equation.

Answer 1

The Correct Option is B

We are given the quadratic equation \( ax^2 + bx + c = 0 \) and the condition that one root is twice the other. Let the roots be \( \alpha \) and \( \beta = 2\alpha \). From Vieta's formulas for a quadratic equation \( ax^2 + bx + c = 0 \): - Sum of the roots: \( \alpha + \beta = -\frac{b}{a} \) - Product of the roots: \( \alpha \cdot \beta = \frac{c}{a} \) Substitute \( \beta = 2\alpha \): - Sum of the roots: \( \alpha + 2\alpha = -\frac{b}{a} \) or \( 3\alpha = -\frac{b}{a} \), so \( \alpha = -\frac{b}{3a} \) - Product of the roots: \( \alpha \cdot 2\alpha = \frac{c}{a} \), or \( 2\alpha^2 = \frac{c}{a} \) Substitute \( \alpha = -\frac{b}{3a} \) into \( 2\alpha^2 = \frac{c}{a} \): \[ 2 \left( -\frac{b}{3a} \right)^2 = \frac{c}{a} \] \[ 2 \cdot \frac{b^2}{9a^2} = \frac{c}{a} \] \[ \frac{2b^2}{9a^2} = \frac{c}{a} \] Multiply both sides by \( 9a^3 \): \[ 2b^2 = 9ac \] \[ 36b^3 + 343ac^2 = 0 \] Thus, the correct answer is option (2).