The number of real values of x for which the equality $| 3 x^{2} + 12 x + 6 | = 5 x + 16$ holds good is

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Solution

The correct option is C
2

Equation $| 3 x^{2} + 12 x + 6 | = 5 x + 16$ ......(i)

when $3 x^{2} + 12 x + 6 \geq 0$ ⇔ $x^{2} + 4 x \geq - 2$

⇔ $| x + 2 |^{2} \geq 4 - 2$ ⇔ $| x + 2 | \geq (\sqrt{2})^{2}$

⇔ x + 2 ≤ - $\sqrt{2}$ or x + 2 ≥ $\sqrt{2}$ .......(ii)

Then (i) becomes $3 x^{2} + 12 x + 6 = 5 x + 16$

⇔ $3 x^{2} + 7 x - 10 = 0$ ⇒ x = 1, - $\frac{10}{3}$

But x = - $\frac{10}{3}$ does not satisfy (ii)

When $3 x^{2} + 12 x + 6 < 0$ ⇒ $x^{2} + 4 x > - 2$

$\Rightarrow | x + 2 | \leq \sqrt{2} \Rightarrow - \sqrt{2} - 2 \leq x \leq - 2 + \sqrt{2}$ ....(iii)

⇒ $3 x^{2} + 17 x + 22 = 0$ ⇒ $x = - 2, - \frac{11}{3}$

But x = - $\frac{11}{3}$ does not satisfy (iii). So, 1 and -2 are the only solutions.

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Similar questions

The number of real values of x for which the equality $| 3 x^{2} + 12 x + 6 | = 5 x + 16$ holds good is

Let , $f (x) = x^{2} + 5 x + 6$ , then the number of real roots of $(f (x))^{2} + 5 f (x) + 6 - x = 0$ is

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\sqrt{3 x^{2} + x + 5} = x - 3

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