Question

If one root of the quadratic equation \( ax^2 + bx + c = 0 \) is double the other, then what is the correct relation among the coefficients?

• A. \( b^2 = 8ac \)
• B. \( b^2 = 4ac \)
• C. \( b^2 = \frac{9ac}{2} \)
• D. \( b^2 = 2ac \)

Hint:

\textbf{Tip:} Assume algebraic relationships for the roots and compare with standard form \( ax^2 + bx + c \) using Vieta's formulas.

Answer 1

The Correct Option is C

To establish the relationship between the coefficients of the quadratic equation \(ax^2+bx+c=0\) when one root is twice the other, perform the following operations:
Let the roots be \(\alpha\) and \(2\alpha\). Applying Vieta's formulas yields:
\( \alpha+2\alpha=-\frac{b}{a} \quad \Rightarrow \quad 3\alpha=-\frac{b}{a} \quad \Rightarrow \quad \alpha=-\frac{b}{3a} \)
Additionally, \( \alpha\cdot2\alpha=\frac{c}{a} \quad \Rightarrow \quad 2\alpha^2=\frac{c}{a} \)
Substitute the expression for \(\alpha\) from \(-\frac{b}{3a}\) into the equation \(2\alpha^2=\frac{c}{a}\):
\(2\left(-\frac{b}{3a}\right)^2=\frac{c}{a}\)
\(2\cdot\frac{b^2}{9a^2}=\frac{c}{a}\)
\(\frac{2b^2}{9a^2}=\frac{c}{a}\)
Eliminate denominators through cross-multiplication:
\(2b^2a=9a^2c \)
After dividing by \(a^2\), the equation simplifies to:
\(b^2=\frac{9ac}{2}\)
Therefore, the governing relation is \(b^2=\frac{9ac}{2}\).