The number of real roots of the equation x | | vedclass

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Question

$x|x|-5|x+2|+6=0$

$C-1:-x \in[0, \infty]$

$x^2-5 x-4=0$

$x=\frac{5 \pm \sqrt{25+16}}{2}=\frac{5+\sqrt{41}}{2}$

$x=\frac{5 \pm \sqrt{41}}{2

Vedclass · Accepted Answer

$3$

Vedclass · Answer

$x|x|-5|x+2|+6=0$

$C-1:-x \in[0, \infty]$

$x^2-5 x-4=0$

$x=\frac{5 \pm \sqrt{25+16}}{2}=\frac{5+\sqrt{41}}{2}$

$x=\frac{5 \pm \sqrt{41}}{2}$

$C-2:-:-x \in[-2,0)$

$-x^2-5 x-4=0$

$x^2+5 x+4=0$

$x=-1,-4$

$x=-1$

$C-3: x \in[-\infty,-2)$

$-x^2+5 x+16=0$

$x^2-5 x-16=0$

$x=\frac{5 \pm \sqrt{25+64}}{2}$

$\frac{5 \pm \sqrt{89}}{2}$

$x=\frac{5-\sqrt{89}}{2}$

The number of real roots of the equation $x | x |-5| x +2|+6=0$, is

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