Solve the system of inequalities: -x^2 + 3x + 4 >= 0 and 3x^2 - 5x >= 0. Give the solution set as a union of intervals.

Step-by-step solution

1. 1

   Treat the word "and" as an instruction to intersect. Each inequality has its own solution set; the system holds exactly on the overlap. So the work splits into two independent quadratic sign analyses plus one intersection at the end.

2. 2

   Flip the first inequality so the parabola opens upward. Multiplying $-x^{2}+3x+4\ge 0$ through by $-1$ reverses the direction:

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

   Since the parabola $y=(x-4)(x+1)$ opens upward, it is at or below zero exactly between its roots:

   $-1\le x\le 4.$

3. 3

   Factor $3x^{2}-5x$ by pulling out $x$. With no constant term the factorisation is immediate:

   $3x^{2}-5x=x(3x-5)\ge 0,\qquad\text{roots }x=0,\;x=\tfrac53 .$

   An upward parabola is $\ge 0$ outside its roots, so

   $x\le 0\quad\text{or}\quad x\ge \tfrac53 .$

   Compare with the companion problem where the factor is $3x-2$: only the right-hand root moves, from $\tfrac23$ to $\tfrac53$.

4. 4

   Intersect $[-1,4]$ with $(-\infty,0]\cup\left[\tfrac53,\infty\right)$. Piece by piece,

   $[-1,4]\cap(-\infty,0]=[-1,0],\qquad[-1,4]\cap\left[\tfrac53,\infty\right)=\left[\tfrac53,4\right],$

   giving the solution set $[-1,0]\cup\left[\tfrac53,4\right]$. The excluded gap $\left(0,\tfrac53\right)$ is wider than in the $3x-2$ version, which is the only visible difference between the two answers.

5. 5

   Spot-check three points. $x=0.5$: first inequality gives $-0.25+1.5+4=5.25\ge 0$ ✓, second gives $0.75-2.5=-1.75<0$ ✗ — correctly excluded. $x=2$: $6\ge 0$ ✓ and $12-10=2\ge 0$ ✓ — included. $x=-1$: $-1-3+4=0\ge 0$ ✓ and $3+5=8\ge 0$ ✓ — the left endpoint belongs to the set.