1. We need to solve $x^2+3x+2=\min\{|x-3|,|x+2|\}.$

First factor the quadratic: $x^2+3x+2=(x+1)(x+2).$

So the equation becomes $(x+1)(x+2)=\min\{|x-3|,|x+2|\}.$

2. Now determine the function $\min\{|x-3|,|x+2|\}.$ This is the smaller of the distances of $x$ from $3$ and from $-2$.

The midpoint of $3$ and $-2$ is $\frac{3+(-2)}{2}=\frac12.$ So:

• if $x\le \frac12$, then $x$ is closer to $-2$, hence $\min\{|x-3|,|x+2|\}=|x+2|;$
• if $x\ge \frac12$, then $x$ is closer to $3$, hence $\min\{|x-3|,|x+2|\}=|x-3|.$

Thus solve piecewise.

3. Case I: $x\le \frac12$

Then the equation is $x^2+3x+2=|x+2|.$ Now split again based on sign of $x+2$.

Case I(a): $x\le -2$

Then $|x+2|=-(x+2)$, so $x^2+3x+2=-x-2$ $x^2+4x+4=0$ $(x+2)^2=0$ $x=-2.$ This satisfies $x\le -2$, so it is a valid solution.

Case I(b): $-2\le x\le \frac12$

Then $|x+2|=x+2$, so $x^2+3x+2=x+2$ $x^2+2x=0$ $x(x+2)=0$ $x=0\quad \text{or}\quad x=-2.$ Both satisfy $-2\le x\le \frac12$, so both are valid.

So from Case I, solutions are $x=-2,\;0.$

4. Case II: $x\ge \frac12$

Then the equation is $x^2+3x+2=|x-3|.$ Again split by sign of $x-3$.

Case II(a): $\frac12\le x\le 3$

Then $|x-3|=3-x$, so $x^2+3x+2=3-x$ $x^2+4x-1=0.$ Solving, $x=\frac{-4\pm\sqrt{16+4}}{2}=\frac{-4\pm\sqrt{20}}{2}=-2\pm\sqrt5.$ Now check interval $\left[\frac12,3\right]$:

• $-2-\sqrt5<0$, invalid
• $-2+\sqrt5\approx 0.236<\frac12$, invalid

So no solution from this subcase.

Case II(b): $x\ge 3$

Then $|x-3|=x-3$, so $x^2+3x+2=x-3$ $x^2+2x+5=0.$ Discriminant: $\Delta=4-20=-16<0,$ so no real solution.

Thus Case II gives no real solution.

5. Therefore the real solutions are $x=-2,\;0.$ So the number of real solutions is $2.$

6. Comparing with the stored correct answer:

• Derived answer: 2
• Stored correct answer: A = 2

They match.