If $p, q, r$ and $s$ are real numbers such that $p r = 2 (q + s)$ , then show that at least one of the equations $x^{2} + p x + q = 0$ and $x^{2} + r x + s = 0$ has real roots.

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Solution

We have $x^{2} + p x + q = 0.... (i)$
$x^{2} + r x + s = 0.... (i i)$

Let $D_{1}$ and $D_{2}$ be the discriminants of equations (i) and (ii). Then

$D_{1} = b^{2} - 4 a c = p^{2} - 4 q$
similarly,
$D_{2} = r^{2} - 4 s$

$\Rightarrow D_{1} + D_{2} = p^{2} - 4 q + r^{2} - 4 s = (p^{2} + r^{2}) - 4 (q + s)$

$\Rightarrow D_{1} + D_{2} = p^{2} + r^{2} - 4 (\frac{p r}{2}) [∵ p r = 2 (q + s) ∴ q + s = \frac{p r}{2}]$

$\Rightarrow D_{1} + D_{2} = p^{2} + r^{2} - 2 p r = {(p - r)}^{2} \geq 0$

At least one of $D_{1}$ and $D_{2}$ is greater than or equal to zero
At least one of the two equations has real roots.

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