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

We must solve the equation $$x\lvert x\rvert \;-\;5\lvert x+2\rvert \;+\;6 \;=\;0$$.
Because of the absolute-value symbols, the algebraic form of the left-hand side changes whenever either $$x=0$$ or $$x=-2$$. Hence analyse the three intervals

Case 1: $$x\lt-2$$

Here $$x$$ is negative and $$x+2$$ is also negative, so $$\lvert x\rvert=-x,\qquad \lvert x+2\rvert=-(x+2).$$ Substituting in the equation gives $$x(-x)\;-\;5\bigl(-(x+2)\bigr)\;+\;6=0$$ $$\Longrightarrow\;-x^{2}+5x+16=0$$ $$\Longrightarrow\;x^{2}-5x-16=0\;-(1)$$

The discriminant of $$(1)$$ is $$25+64=89,$$ so the roots are $$\dfrac{5\pm\sqrt{89}}{2}.$$ The negative root is $$\dfrac{5-\sqrt{89}}{2}\approx-2.217,$$ which lies in the interval $$(-\infty,-2).$$ The positive root $$\dfrac{5+\sqrt{89}}{2}\approx7.217$$ does not belong to this interval.
Hence Case 1 contributes exactly one real root.

Case 2: $$-2\le x&lt0$$

Now $$x$$ is still negative whereas $$x+2$$ is non-negative, so $$\lvert x\rvert=-x,\qquad \lvert x+2\rvert=x+2.$$ Substituting, $$x(-x)\;-\;5(x+2)\;+\;6=0$$ $$\Longrightarrow\;-x^{2}-5x-4=0$$ $$\Longrightarrow\;x^{2}+5x+4=0\;-(2)$$

The discriminant of $$(2)$$ is $$25-16=9,$$ giving roots $$x=-1,\;x=-4.$$ Only $$x=-1$$ lies in the interval $$[-2,0),$$ so Case 2 contributes one real root.

Case 3: $$x\ge0$$

Here both $$x$$ and $$x+2$$ are non-negative, so $$\lvert x\rvert=x,\qquad \lvert x+2\rvert=x+2.$$ Thus $$x(x)\;-\;5(x+2)\;+\;6=0$$ $$\Longrightarrow\;x^{2}-5x-4=0\;-(3)$$

The discriminant of $$(3)$$ is $$25+16=41,$$ so the roots are $$x=\dfrac{5\pm\sqrt{41}}{2}.$$ The positive root $$\dfrac{5+\sqrt{41}}{2}\approx5.701$$ belongs to $$[0,\infty)$$, whereas the negative root $$\dfrac{5-\sqrt{41}}{2}\approx-0.701$$ does not. Therefore Case 3 adds one more real root.

Summing up:

• Case 1: one root (approximately $$-2.217$$)
• Case 2: one root (exactly $$-1$$)
• Case 3: one root (approximately $$5.701$$)

The original equation has $$1+1+1=3$$ real solutions.

Hence the number of real roots is $$3$$. Therefore Option D is correct.