Find the set of values of $^{'} p^{'}$ for which the quadratic equation, $(p - 5) x^{2} - 2 p x - 4 p = 0$ has real roots.

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Solution

$\begin{matrix} (p - 5) x^{2} - 2 p x - 4 p = 0 D \geq 0 (- 2 p^{2}) + 4 (p - 5) \times 4 p \geq 0 4 p^{2} + 16 p (p - 5) \geq 0 4 p (p + 4 (p - 5)) \geq 0 4 p {p + 4 p - 20} \geq 0 4 p (5 p - 20) \geq 0 p \geq 0 a n d p \leq 4 p \in [0, 4] \end{matrix}$

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