If pr = 2 (q + s) then among the equations x2+px+q = 0 and x2 + rx + s = 0

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If pr = 2 (q + s) then among the equations x²+px+q = 0 and x² + rx + s = 0

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at least one has real roots

at least one has imaginary roots

both have imaginary roots

both have real roots

answer is C.

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Detailed Solution

$\begin{aligned} Δ_{1} + Δ_{2} & = p^{2} + r^{2} - 4 (q + s) = p^{2} + r^{2} - 2 pr \\ = (p - r)^{2} \geq 0 \Rightarrow at least one is + ve \end{aligned}$

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