lf $a, b, c, d$ are real and no two of them are simultaneously zero, then the equation $(x^{2} + a x - 3 b) (x^{2} - c x + b) (x^{2} - d x + 2 b) = 0$ has

at least two real roots

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at least four real roots

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all roots real

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at least two imaginary roots

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Solution

The correct option is A at least two real roots
Let $D_{1}, D_{2}, D_{3}$ be discriminants of
$x^{2} + a x - 3 b = 0$ ,
$x^{2} - c x + b = 0$ ,
$x^{2} - d x + 2 b = 0$ respectively.

$D_{1} = a^{2} + 12 b$
$D_{2} = c^{2} - 4 b$
$D_{3} = d^{2} - 8 b$
$\Rightarrow D_{1} + D_{2} + D_{3} = a^{2} + 12 b + c^{2} - 4 b + d^{2} - 8 b$ $= a^{2} + c^{2} + d^{2}$
$\Rightarrow D_{1} + D_{2} + D_{3} > 0$
At least one of $D_{1}, D_{2}, D_{3}$ is greater than zero .
$\Rightarrow$ The equation has at least $2$ real roots.

Hence, option A.

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Similar questions

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(x^{2} + a x - 3 b) (x^{2} - c x + b) (x^{2} - d x + 2 b) = 0

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