Sum of the roots of the equation $(x + 3)^{2} - 4 | x + 3 | + 3 = 0$ is-

4

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12

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- 12

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- 4

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Solution

The correct option is C $- 12$
${(x + 3)}^{2} - 4 | x + 3 | + 3 = 0 l e t | x + 3 | = t ∴ t \geq 0 t^{2} - 4 t + 3 = 0 (t - 1) (t - 3) = 0 ∴ t = 1 o r 3 w h e n t = 1 | x + 3 | = 1 x + 3 = 1 o r x + 3 = - 1 x = - 2 o r x = - 4 w h e n t = 3 | x + 3 | = 3 o r x + 3 = - 3 x = 0 o r x = - 6 s u m o f a l l r o o t s = (- 2) + (- 4) + (0) + (- 6) = - 12$

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