Find the value of $p$
$2 x^{2} + 3 x + p = 0$ has real roots.

p \geq \frac{1}{8}

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p \leq \frac{9}{8}

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p > \frac{9}{8}

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p < \frac{1}{8}

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Solution

The correct option is A $p \leq \frac{9}{8}$
For real roots, we have $D ⩾ 0$
$\Rightarrow$ $(3)^{2} - 4 (2) (p) ⩾ 0$
$\Rightarrow$ $9 ⩾ 8 p$
$\Rightarrow$ $\frac{9}{8} ⩾ p$

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