The number of distinct real solutions of the equation $$x\lvert x+4 \rvert + 3\lvert x+2 \rvert + 10 = 0$$ is

The equation given to us is $$x\lvert x+4 \rvert + 3\lvert x+2 \rvert + 10 = 0$$

Critical points for this are $$ x = -4, -2$$

Case 1: $$ x \ge -2$$

Then, $$ |x+4| = x+4,\quad |x+2| = x+2 $$

The equation becomes, 

$$ x(x+4) + 3(x+2) + 10 = 0$$
or, $$ x^2 + 4x + 3x + 6 + 10 = 0$$
or, $$  x^2 + 7x + 16 = 0 $$

For this equation, the discriminant is 49 - 64 = -15 < 0. So, this has no real solution.

Case 2: $$ -4 \le x < -2$$

Then, $$ |x+4| = x+4,\quad |x+2| = -(x+2) $$

The equation becomes,

$$ x(x+4) + 3(-(x+2)) + 10 = 0$$
or, $$ x^2 + 4x - 3x - 6 + 10 = 0 $$
or, $$ x^2 + x + 4 = 0 $$

For this equation too, the discriminant is  1 - 16 = -15 < 0. So this too has no real solution.

Case 3: $$ x < -4$$

Then, $$ |x+4| = -(x+4),\quad |x+2| = -(x+2) $$

The equation becomes, 

$$x(-(x+4)) + 3(-(x+2)) + 10 = 0$$
or, $$ x^2 - 4x - 3x - 6 + 10 = 0 $$
or, $$ x^2 - 7x + 4 = 0 $$

or, $$ x = \dfrac{-7 \pm \sqrt{49 + 16}}{2} = \dfrac{-7 \pm \sqrt{65}}{2} $$

Now we check if the solution values lie in the domain x < -4:

• $$ \frac{-7 + \sqrt{65}}{2} \approx 0.53$$ , which is not a valid solution
• $$ \frac{-7 - \sqrt{65}}{2} \approx -7.53$$ , which is a valid solution. 

Hence, the number of distinct real solutions is 1.