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Discriminant

If at least o...

Question

If at least one of the equations $x^{2} + p x + q = 0$ , $x^{2} + r x + s = 0$ has real roots, then

A

q s = (p + r)

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B

p r = (q + s)

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C

p r = 2 (q + s)

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D

None of these.

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Solution

The correct option is D $p r = 2 (q + s)$
Given at least one of the equation $x^{2} + p x + q = 0$ , $x^{2} + r x + s = 0$ has real roots.
Let $D_{1}$ and $D_{2}$ be the discriminant values.
At least one of the equation has real roots when $D_{1} + D_{2} \geq 0$ .
$\Rightarrow p^{2} - 4 q + r^{2} - 4 s \geq 0$
$\Rightarrow (p - r)^{2} + 2 p r - 4 (q + s) \geq 0$
Above inequality holds only if $2 p r - 4 (q + s) = 0$ .
$∴ p r = 2 (q + s)$ .
Hence, option C.

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