Number of real roots of equation $x^{4} + x^{2} - 12 = 0$ is:

4

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2

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0

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1

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Solution

The correct option is B $2$
$x^{4} + x^{2} - 12 = 0$

roots of the equation are $\pm \sqrt{3}, \pm 2 i$

So Number of real roots of eq is 2.

$x^{4} + x^{2} - 12 = 0$

$= (x^{2} - 3) (x^{2} + 4) = 0$

$x = \pm \sqrt{3}$ $x = \pm 2 i$

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Similar questions

x^{4} - x^{2} + 2 x - 1 = 0

f (x) = 1 + \frac{x}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \frac{x^{4}}{4!}

f (x) = 0

x^{4} + x^{2} + 1 = 0

x^{3} - x^{2} + x - 1 = 0

x^{4} - 1 = 0

x^{4} + 4 x^{3} - 2 x^{2} - 12 x + k = 0

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