If equation ${p x}^{2} + q x + r = 0$ has real roots, then $r = ?$

r \leq \frac{4 p}{q^{2}}

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r \geq \frac{q^{2}}{4 p}

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r = \frac{p}{q^{2}}

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r = \frac{{4 q}^{2}}{p}

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Solution

The correct option is A $r \geq \frac{q^{2}}{4 p}$
Given equation is $p x^{2} + q x + r = 0$
For real roots $Δ \geq 0$ i.e. $b^{2} - 4 a c \geq 0$
Here $a = p, b = q, c = r$
Therefore, $q^{2} - 4 p r \geq 0$
$\Rightarrow q^{2} \geq 4 p r$
$\Rightarrow \frac{q^{2}}{4 p} \geq r, r \leq \frac{q^{2}}{4 p}$

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