Let the set of all values of , for which both the roots of the equation are negative real numbers, be the interval . Then is equal to [2025]

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Solution

Q.
Let the set of all values of $p \in R$ , for which both the roots of the equation $x^{2} - (p + 2) x + (2 p + 9) = 0$ are negative real numbers, be the interval $(α, β]$ . Then $β - 2 α$ is equal to [2025]

1 5

2 0

3 20

4 9

Ans.
(1)

Given, $x^{2} - (p + 2) x + (2 p + 9) = 0$

$D \geq 0$

$\Rightarrow {(p + 2)}^{2} - 4 (2 p + 9) \geq 0$

$\Rightarrow p^{2} + 4 p + 4 - 8 p - 36 \geq 0$

$\Rightarrow p^{2} - 4 p - 32 \geq 0$

$\Rightarrow (p - 8) (p + 4) \geq 0$

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

So, sum of roots : p + 2 < 0

$\Rightarrow$ p < –2 ... (ii)

and product of roots : 2p + 9 > 0

$\Rightarrow p > \frac{- 9}{2}$ ... (iii)

Using (i), (ii) and (iii), we get

$p \in (\frac{- 9}{2}, - 4]$

$∴ β - 2 α = - 4 - 2 (\frac{- 9}{2}) = 5$ .

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