ANSWER
C. x² - x ≥ 6
EXPLANATION
Let's analyze the solution set graphed first. We can see that the values -2 and 3 are included in the set, and all values below -2 and above 3. So, the solution set is (-∞, 2] U [3, ∞).
To find which inequality satisfies this solution set we have to solve them. To do so, we will be using the quadratic formula:
[tex]\begin{gathered} ax^2+bx+c=0 \\ \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{gathered}[/tex]A. To solve this one, first, add x to both sides,
[tex]-x^2+x+6\geqslant0[/tex]Now, apply the quadratic formula to find the zeros. For this inequality, a = -1, b = 1, and c = 6
[tex]\begin{gathered} x=\frac{-1\pm\sqrt{1^2-4(-1)6}}{2(-1)}=\frac{-1\pm\sqrt{1+24}}{-2}=\frac{-1\pm\sqrt{25}}{-2} \\ \\ x_1=\frac{-1-5}{-2}=\frac{-6}{-2}=3 \\ \\ x_2=\frac{-1+5}{-2}=\frac{4}{-2}=-2 \end{gathered}[/tex]But in this case, the solution set is [-2, 3] - note that for any value outside this interval the inequality is false.
B. Similarly, apply the quadratic formula for a = -3, b = 3, c = 18,
[tex]\begin{gathered} x=\frac{-3\pm\sqrt{3^2-4(-3)18}}{2(-3)}=\frac{-3\pm\sqrt{9+216}}{2(-3)}=\frac{-3\pm\sqrt{225}}{-6}=\frac{-3\pm15}{-6} \\ \\ x_1=\frac{-3+15}{-6}=\frac{12}{-6}=-2 \\ \\ x_2=\frac{-3-15}{-6}=\frac{-18}{-6}=3 \end{gathered}[/tex]Again, the solution set is [-2, 3] since for any value outside the interval the inequality is not true.
C. Subtract 6 from both sides,
[tex]x^2-x-6\geqslant0[/tex]Apply the quadratic formula, with a = 1, b = -1, and c = -6,
[tex]\begin{gathered} x=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot1(-6)}}{2\cdot1}=\frac{1\pm\sqrt{1+24}}{2}=\frac{1\pm\sqrt{25}}{2}=\frac{1\pm5}{2} \\ \\ x_1=\frac{1+5}{2}=\frac{6}{2}=3 \\ \\ x_2=\frac{1-5}{2}=\frac{-4}{2}=-2 \end{gathered}[/tex]In this case, if we take any value between -2 and 3, for example 1,
[tex]\begin{gathered} 1^2-1\ge6 \\ \\ 0\ge6 \end{gathered}[/tex]We can see that the inequality is false, while if we take a value greater than 3 or less than -2, for example, -5,
[tex]\begin{gathered} (-5)^2-(-5)\ge6 \\ \\ 25+5\ge6 \\ \\ 30\ge6 \end{gathered}[/tex]We can see that the inequality is true.
Hence, we can conclude that inequality C satisfies the solution set graphed.
