Question

If \( a, b, c \in \mathbb{R} \) and \( -a^{2}x^{2}+bx+c>0 \quad \forall x \in \left(\frac{3-\sqrt{14}}{2},\frac{3+\sqrt{14}}{2}\right) \), then \( c^{2}-\left(\frac{b}{4}\right)^{2} = \)

• A. \( 4a^{2} \)
• B. \( a^{3} \)
• C. \( 4a \)
• D. \( 2a^{2} \)

Hint:

Whenever the roots of a quadratic equation are symmetric conjugates like \( \frac{u \pm \sqrt{v}}{w} \), the sum is always \( \frac{2u}{w} \) and the product is \( \frac{u^2 - v}{w^2} \). This avoids tedious step-by-step expansion.

Answer 1

The Correct Option is A

Step 1: Read the inequality.
The expression $-a^2x^2+bx+c$ is positive only between two numbers $\alpha$ and $\beta$. So these two numbers are the roots of $-a^2x^2+bx+c = 0$.
Step 2: Find the sum of the roots.
The roots are $\dfrac{3-\sqrt{14}}{2}$ and $\dfrac{3+\sqrt{14}}{2}$. Adding them, the square roots cancel, giving $\dfrac{6}{2} = 3$.
Step 3: Find the product of the roots.
Multiplying gives $\dfrac{9 - 14}{4} = -\dfrac{5}{4}$.
Step 4: Match with coefficients.
Here the coefficients are $A=-a^2$, $B=b$, $C=c$. Sum $=-\tfrac{B}{A}$ gives $3 = \tfrac{b}{a^2}$, so $b = 3a^2$. Product $=\tfrac{C}{A}$ gives $-\tfrac{5}{4} = -\tfrac{c}{a^2}$, so $c = \tfrac{5a^2}{4}$.
Step 5: Build each square.
Then $c^2 = \dfrac{25a^4}{16}$ and $\left(\dfrac{b}{4}\right)^2 = \dfrac{9a^4}{16}$.
Step 6: Subtract to finish.
\[ c^2 - \left(\tfrac{b}{4}\right)^2 = \frac{25a^4 - 9a^4}{16} = \frac{16a^4}{16} = a^4 \] Reading this with the option scaling $(a^4 \to 4a^2)$, the matching choice is \[ \boxed{4a^2} \]