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 determine the correct relation among the coefficients of the quadratic equation \(ax^2+bx+c=0\) when one root is double the other, follow these steps: 
Let the roots be \(\alpha\) and \(2\alpha\). By Vieta's formulas, we know:
\( \alpha+2\alpha=-\frac{b}{a} \quad \Rightarrow \quad 3\alpha=-\frac{b}{a} \quad \Rightarrow \quad \alpha=-\frac{b}{3a} \)
Also, \( \alpha\cdot2\alpha=\frac{c}{a} \quad \Rightarrow \quad 2\alpha^2=\frac{c}{a} \)
Substitute \(\alpha=-\frac{b}{3a}\) into \(2\alpha^2=\frac{c}{a}\):
\(2\left(-\frac{b}{3a}\right)^2=\frac{c}{a}\)
\(2\cdot\frac{b^2}{9a^2}=\frac{c}{a}\)
\(\frac{2b^2}{9a^2}=\frac{c}{a}\)
Cross-multiply to clear fractions:
\(2b^2a=9a^2c \)
Dividing by \(a^2\), we get:
\(b^2=\frac{9ac}{2}\)
Thus, the correct relation is \(b^2=\frac{9ac}{2}\).

Answer 2

To solve this problem, we need to analyze the condition given: one root of the quadratic equation \(ax^2 + bx + c = 0\) is double the other. Let the roots be \(r\) and \(2r\).

Using Vieta's formulas, for a quadratic equation \(ax^2+bx+c=0\), the sum of the roots is given by:

\[r + 2r = -\frac{b}{a}\]

This simplifies to:

\[3r = -\frac{b}{a}\]

Thus,

\[r = -\frac{b}{3a}\]

The product of the roots is given by:

\[r \cdot 2r = \frac{c}{a}\]

This simplifies to:

\[2r^2 = \frac{c}{a}\]

Substitute \(r = -\frac{b}{3a}\) into the product of roots expression:

\[2\left(-\frac{b}{3a}\right)^2 = \frac{c}{a}\]

Which simplifies to:

\[2\left(\frac{b^2}{9a^2}\right) = \frac{c}{a}\]

Thus,

\[\frac{2b^2}{9a^2} = \frac{c}{a}\]

By cross-multiplying, we obtain:

\[2b^2 = 9ac\]

Therefore, the correct relation among the coefficients is:

\[b^2 = \frac{9ac}{2}\]