Solve for $x : 3 x^{2} - 12 = x (1 + 2 x)$

x = 4

- 4

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x = 3

- 3

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x = 4

- 3

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None of these

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Solution

The correct option is D $x = 4$ or $- 3$
Given, $3 x^{2} - 12 = x (1 + 2 x)$
$\Rightarrow 3 x^{2} - 12 = x + 2 x^{2}$
$\Rightarrow x^{2} - x - 12 = 0$
$\Rightarrow x^{2} - 4 x + 3 x - 12 = 0$
We get $x (x - 4) + 3 (x - 4) = 0$
$\Rightarrow (x - 4) (x + 3) = 0$
Therefore $x = 4, - 3$

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