JEE Mathematics 2025 Question with Solution

Given: The quadratic equation is $x^2 - (p + 2)x + (2p + 9) = 0$.

Find: The value of $\beta - 2\alpha$ when both roots are negative real numbers and the set of such values of $p$ is $(\alpha, \beta)$.

For both roots to be negative real numbers, we use the conditions on sum, product, and discriminant.

1. Sum of roots:

$$\alpha_1 + \alpha_2 = p + 2$$

For both roots to be negative,

$$p + 2 < 0$$

So,

$$p < -2$$

2. Product of roots:

$$\alpha_1 \alpha_2 = 2p + 9$$

For both roots to be negative, their product must be positive.

$$2p + 9 > 0$$

So,

$$p > -\frac{9}{2}$$

3. Discriminant condition for real roots:

$$D = (p + 2)^2 - 4(2p + 9) \ge 0$$ $$p^2 + 4p + 4 - 8p - 36 \ge 0$$ $$p^2 - 4p - 32 \ge 0$$ $$(p - 8)(p + 4) \ge 0$$

Hence,

$$p \in (-\infty, -4] \cup [8, \infty)$$

Combining all conditions:

$$p < -2, \quad p > -\frac{9}{2}, \quad p \in (-\infty, -4] \cup [8, \infty)$$

We get,

$$p \in \left[-\frac{9}{2}, -4\right]$$

Thus,

$$\alpha = -\frac{9}{2}, \quad \beta = -4$$

Now,

$$\beta - 2\alpha = -4 - 2\left(-\frac{9}{2}\right)$$ $$= -4 + 9 = 5$$

Therefore, $\beta - 2\alpha = 5$ and the correct option is C.](streamdown:incomplete-link)