Quadratic Equations
Negative roots with parameter
nta_pyq_2025_apr
Grade 12
Question:
Let the set of all values of $p \in \mathbb{R}$, for which both the roots of the equation $x^2 - (p + 2)x + (2p + 9) = 0$ are negative real numbers, be the interval $(\alpha, \beta]$. Then $\beta - 2\alpha$ is equal to
Step-by-Step Solution
Key Concept: Use the location of roots for a quadratic to force both roots negative: real roots, negative sum, and positive product.
Using location of roots:
(i) $\Delta \geq 0$
(ii) $-\frac{b}{2a} < 0$
(iii) $a \cdot f(0) > 0$
$[p + 2]^2 - 4(2p + 9) \geq 0$
$(p + 4)(p - 8) \geq 0$
$p + 2 < 0$
$2p + 9 > 0$
Intersection: $p \in \left(-\frac{9}{2}, -4\right]$
$\therefore \alpha = -\frac{9}{2}$, $\beta = -4$
$\beta - 2\alpha = -4 - 2\left(-\frac{9}{2}\right) = -4 + 9 = 5$
Correct Answer: 3