Sum of real roots of $x^{2} + | x | - 6 = 0$ is

4

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- 6

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- 1

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0

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Solution

The correct option is D $0$
When $x \geq 0$
$x^{2} + x - 6 = 0$
$(x + 3) (x - 2) = 0$
$x = - 3$ & $2$
Since $x \geq 0$ , therefore $x = 2$
When $x < 0$
$x^{2} - x - 6 = 0$
$(x - 3) (x + 2) = 0$
$x = 3$ & $- 2$
Since $x < 0$ , therefore $x = - 2$
Therefore sum of roots $2 - 2 = 0$

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