-12greater than or equal to x^(2)+8x
x >= 2
How do you get that
-12 >= x^2+8x
x^2+8x+12 <= 0
(x+2)(x+6) <= 0
You know the roots are -2 and -6, and the parabola opens upward. So, y is negative between the roots:
-6 <= x <= -2
Clearly Philip did not check his answer.
To solve this inequality, we first need to isolate the variable x.
So, let's rearrange the inequality:
x^2 + 8x + 12 ≤ 0
Now, we aim to factorize the quadratic equation to identify the solutions.
Let's try to find two numbers, a and b, which have the following properties:
- Their product is equal to 12 (coefficient of x^2 term multiplied by the constant term),
- Their sum is equal to 8 (coefficient of x term).
By matching these properties, we can write the quadratic equation as a product of two linear binomials:
(x + a)(x + b) ≤ 0
Now, we need to find a and b such that (x + a)(x + b) ≤ 0.
The factors of 12 are: ±1, ±2, ±3, ±4, ±6, and ±12.
We need to find two numbers, a and b, that have different signs. Not all combinations will give us what we need, so we need to test them.
After checking the combinations, we find that the pairs (a, b) satisfying the conditions are: (1, 12), (-1, -12), (2, 6), (-2, -6), (3, 4), (-3, -4).
For each pair of factors, we can write two inequalities:
1. (x + 1)(x + 12) ≤ 0
2. (x - 1)(x - 12) ≤ 0
3. (x + 2)(x + 6) ≤ 0
4. (x - 2)(x - 6) ≤ 0
5. (x + 3)(x + 4) ≤ 0
6. (x - 3)(x - 4) ≤ 0
Now, we need to solve each of these inequalities separately.
For example, let's solve the first inequality: (x + 1)(x + 12) ≤ 0.
To solve this inequality, we set each factor equal to zero and find the critical points:
x + 1 = 0 => x = -1
x + 12 = 0 => x = -12
These critical points divide the number line into three regions: x < -12, -12 < x < -1, and x > -1.
Next, we choose test points from each region and substitute them into the inequality to determine the sign of the expression:
For the region x < -12, we can choose x = -13.
(-13 + 1)(-13 + 12) = (-12)(-1) = 12, which is greater than zero. So, this region is not a solution.
For the region -12 < x < -1, we can choose x = -2.
(-2 + 1)(-2 + 12) = (-1)(10) = -10, which is less than zero. So, this region is a solution.
For the region x > -1, we can choose x = 0.
(0 + 1)(0 + 12) = (1)(12) = 12, which is greater than zero. So, this region is not a solution.
Therefore, the solution for the inequality (x + 1)(x + 12) ≤ 0 is -12 ≤ x ≤ -1.
Similarly, you can solve the remaining five inequalities, and the final solution will be the union of all the individual solutions.
I hope this explanation helps! Let me know if you have any further questions.