The product of non-real roots of the equation $(x^{2} + x - 2) (x^{2} + x - 3) = 12$ is

1

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2

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3

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Solution

The correct option is A $1$
Let $x^{2} + x - 2 = t$
$\Rightarrow t (t - 1) = 12 \Rightarrow t = 4, - 3$

$∴ x^{2} + x - 2 = 4$ or $x^{2} + x - 2 = - 3$
$\Rightarrow x^{2} + x - 6 = 0$ or $x^{2} + x + 1 = 0$
$x^{2} + x + 1 = 0$ has non-real roots.
$∴$ Product of non-real roots $= 1$

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Similar questions

The number of real roots of the equation $(x^{2} + 2 x)^{2} - (x + 1)^{2} - 55 = 0$

(x^{2} + 2 x)^{2} - (x + 1)^{2} - 55 = 0

(x^{2} + x - 2) (x^{2} + x - 3) = 12

(x^{2} + x - 2) (x^{2} + x - 3) = 12

(x^{2} + x - 2) (x^{2} + x - 3) = 12

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