The possible values of p for which the equation $x^{2} + p x + 64 = 0$ and $x^{2} - 8 x + p = 0$ will both have real roots is

16

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8

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4

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12

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Solution

The correct option is D $16$
$Δ \geq 0$ for both equations
$x^{2} - 8 x + p = 0, x^{2} + p x + 64 = 0$
$64 - 4 p \geq 0, p^{2} - 256 \geq 0$
$16 p \geq 0, p^{2} \geq 256$
$p \leq 16, p \geq \pm 16$

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p > 0

x^{2} + p x + 64 = 0

x^{2} - 8 x + p = 0

x^{2} + 2 p x + 64 = 0 a n d X^{2} - 8 x + 2 p = 0

K

x^{2} + k x + 64 = 0

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λ

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