The number of real values of x for which the | vedclass

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Question

(c) Equation is $|3{x^2} + 12x + 6|\, = 5x + 16$ &hellip;..$(i)$

when $3{x^2} + 12x + 6 \ge 0$$ \Leftrightarrow \,\,\,{x^2} + 4x \ge - 2$

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Vedclass · Accepted Answer

$2$

Vedclass · Answer

(c) Equation is $|3{x^2} + 12x + 6|\, = 5x + 16$ &hellip;..$(i)$

when $3{x^2} + 12x + 6 \ge 0$$ \Leftrightarrow \,\,\,{x^2} + 4x \ge - 2$

$ \Leftrightarrow \,\,\,|x + 2{|^2} \ge 4 - 2\,\,\, \Leftrightarrow \,\,\,|x + 2|\, \ge {(\sqrt 2 )^2}$

$ \Leftrightarrow \,\,\,x + 2 \le - \sqrt 2 $ or $x + 2 \ge \sqrt 2 $ ....$(ii)$

Then $(i)$ becomes $3{x^2} + 12x + 6 = 5x + 16$

$ \Leftrightarrow \,\,\,3{x^2} + 7x - 10 = 0 \Rightarrow x = 1,\, - \frac{{10}}{3}$

But $x = - \frac{{10}}{3}$ does not satisfy $(ii).$

When $3{x^2} + 12x + 6 < 0$$ \Rightarrow {x^2} + 4x < - 2$

== > $\,|x + 2|\, \le \sqrt 2 $$ \Rightarrow \,\, - \sqrt 2 - 2 \le x \le - 2 + \sqrt 2 $ &hellip;..$(iii)$

Then $(i)$ becomes $ \Rightarrow 3{x^2} + 12x + 6 = - (5x + 16)$

$ \Rightarrow \,\,\,\,3{x^2} + 17x + 22 = 0 \Rightarrow x = - 2, - \frac{{11}}{3}$

But $x = - \frac{{11}}{3}$does not satisfy $(iii)$. So, $1$ and $-2$ are the only solutions.

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

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