If both roots of the quadratic equation $(2 - x) (x + 1) = p$ are distinct & positive then find the value of $p$ .

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Solution

Given,

$(2 - x) (x + 1) = p$

$- x^{2} + x + 2 - p = 0$

condition for quadratic equation to have distinct and positive roots is:

$b^{2} - 4 a c \geq 0$

$b^{2} \geq 4 a c$

Here, $a = - 1, b = 1, c = 2 - p$

$(1)^{2} \geq 4 (- 1) (2 - p)$

$1 \geq - 4 (2 - p)$

$1 \geq - 8 + 4 p$

$∴ p \leq \frac{9}{4}$

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