Questions: The inequality x^2 + 12x + 35 ≥ 0 has two critical points and three possible intervals for solutions. Choose each set of possible test points for the three intervals. -8, -6, -4 -10, -6, 0 -6, 0, 6 -6, 0, 10

Solution Steps

To solve the inequality \(x^2 + 12x + 35 \geq 0\), we first need to find the critical points by solving the equation \(x^2 + 12x + 35 = 0\). These critical points will divide the number line into intervals. We then choose test points from each interval to determine where the inequality holds true. The correct set of test points will include one point from each interval.

To solve the inequality \(x^2 + 12x + 35 \geq 0\), we first find the critical points by solving the equation \(x^2 + 12x + 35 = 0\). Factoring the quadratic equation, we get:

\[ (x + 7)(x + 5) = 0 \]

Thus, the critical points are \(x = -7\) and \(x = -5\).

The critical points divide the number line into three intervals: \((-\infty, -7)\), \((-7, -5)\), and \((-5, \infty)\).

We need to choose test points from each interval to determine where the inequality \(x^2 + 12x + 35 \geq 0\) holds true. The possible sets of test points given are:

• \([-8, -6, -4]\)
• \([-10, -6, 0]\)
• \([-6, 0, 6]\)
• \([-6, 0, 10]\)

We evaluate the inequality at each test point in the sets:

1. For the set \([-8, -6, -4]\):

   • At \(x = -8\), \(x^2 + 12x + 35 = 9 \geq 0\)
   • At \(x = -6\), \(x^2 + 12x + 35 = 7 \geq 0\)
   • At \(x = -4\), \(x^2 + 12x + 35 = 3 \geq 0\)

2. For the set \([-10, -6, 0]\):

   • At \(x = -10\), \(x^2 + 12x + 35 = 15 \geq 0\)
   • At \(x = -6\), \(x^2 + 12x + 35 = 7 \geq 0\)
   • At \(x = 0\), \(x^2 + 12x + 35 = 35 \geq 0\)

3. For the set \([-6, 0, 6]\):

   • At \(x = -6\), \(x^2 + 12x + 35 = 7 \geq 0\)
   • At \(x = 0\), \(x^2 + 12x + 35 = 35 \geq 0\)
   • At \(x = 6\), \(x^2 + 12x + 35 = 107 \geq 0\)

4. For the set \([-6, 0, 10]\):

   • At \(x = -6\), \(x^2 + 12x + 35 = 7 \geq 0\)
   • At \(x = 0\), \(x^2 + 12x + 35 = 35 \geq 0\)
   • At \(x = 10\), \(x^2 + 12x + 35 = 255 \geq 0\)

Final Answer