Solve the system of inequalities: -x^2 + 3x + 4 >= 0 and 3x^2 - 2x >= 0. Give the solution set as a union of intervals.

Step-by-step solution

1. 1

   Plan: solve each inequality alone, then intersect. A system joined by and is satisfied only where both statements hold, so the answer is the intersection of two solution sets. Solving them together in one line is impossible; the reliable method is two separate sign analyses followed by one number line.

2. 2

   Make the leading coefficient positive in the first inequality. Multiplying $-x^{2}+3x+4\ge 0$ by $-1$ reverses the inequality sign:

   $x^{2}-3x-4\le 0\;\Longrightarrow\;(x-4)(x+1)\le 0.$

   A product of two factors is $\le 0$ between the roots, and an upward-opening parabola sits below the axis between its roots, so

   $-1\le x\le 4.$

3. 3

   Factor the second inequality by taking out the common factor. There is no constant term, so factor $x$ out rather than reaching for the quadratic formula:

   $3x^{2}-2x=x(3x-2)\ge 0,$

   with roots $x=0$ and $x=\tfrac23$. This parabola also opens upward, so it is $\ge 0$ outside the roots:

   $x\le 0\quad\text{or}\quad x\ge \tfrac23.$

   Note the direction is the opposite of step 2 — that is the whole difference between $\le$ and $\ge$ for an upward parabola.

4. 4

   Intersect the two sets on one number line. The first set is $[-1,4]$; the second is $(-\infty,0]\cup[\tfrac23,\infty)$. Cutting the first against each piece of the second:

   $[-1,4]\cap(-\infty,0]=[-1,0],\qquad [-1,4]\cap\left[\tfrac23,\infty\right)=\left[\tfrac23,4\right].$

   So the solution is $[-1,0]\cup\left[\tfrac23,4\right]$. The gap $\left(0,\tfrac23\right)$ is excluded because there $x(3x-2)<0$.

5. 5

   Check the four endpoints and one point in every region. At $x=0.3$ (inside the gap): $-0.09+0.9+4=4.81\ge0$ ✓ but $3(0.09)-0.6=-0.33<0$ ✗, so $0.3$ is correctly excluded. At $x=-0.5$: $-0.25-1.5+4=2.25\ge0$ ✓ and $0.75+1=1.75\ge0$ ✓. At $x=2$: $6\ge0$ ✓ and $12-4=8\ge0$ ✓. At $x=4.5$: $-20.25+13.5+4<0$ ✗. All four endpoints give equality in one factor, so they belong to the closed set.