The number of real solutions of the equation | vedclass

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Question

Case $I$ : $x \geq-5$

$ x^2+5 x+2 x+12=0 $

$ x^2+7 x+12=0 $

$ x=-3,-4$

Case $II$ : $-7<\mathrm{x}<-5$

$-x^2-5 x+2 x+14-2=0$

$

Vedclass · Accepted Answer

$3$

Vedclass · Answer

Case $I$ : $x \geq-5$

$ x^2+5 x+2 x+12=0 $

$ x^2+7 x+12=0 $

$ x=-3,-4$

Case $II$ : $-7<\mathrm{x}<-5$

$-x^2-5 x+2 x+14-2=0$

$ -x^2-3 x+12=0 $

$ x=\frac{-3 \pm \sqrt{9+48}}{2} $

$ =\frac{-3 \pm \sqrt{57}}{2} $

$ x=\frac{-3-\sqrt{57}}{2}, \frac{-3+\sqrt{57}}{2} 	ext { (rejected) }$

Case $III$ : $\mathrm{x} \leq-7$

$ -x^2-5 x-2 x-14-2=0 $

$ x^2+7 x+16=0 $

$ D=49-64<0$

No solutions

No. of solutions $=3$

The number of real solutions of the equation $\mathrm{x}|\mathrm{x}+5|+2|\mathrm{x}+7|-2=0$ is .....................

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