i need help solving these inequalities i'm not sure if i should factor things out or not
4x^3 is greater than or equal to 16x
and
4 is greater than or equal to 2x^3 + 8x
Yes, you can factor out numbers which are positive as in equations.
4x>16 can be written as
x>4
(common factor 4 is positive.)
If the factor is negative, you can still factor it out, but you will need to reverse the sign of inequality.
-5x ≥ 40
x ≤ 40/(-5)
x ≤ -8
(common factor -5 is negative)
If the sign of the factor is unknown, then you'll have to make different cases:
ax > ab
x > b (if a>0)
x < b (if a<0)
x is undefined if a=0
-y-4+2y>11
To solve these inequalities, you can use algebraic manipulations to isolate the variable on one side of the inequality sign. Let's go through each inequality step by step:
1. 4x^3 ≥ 16x
Start by subtracting 16x from both sides of the inequality to move all terms to one side:
4x^3 - 16x ≥ 0
Next, factor out 4x from each term:
4x(x^2 - 4) ≥ 0
Now, we have a trinomial x^2 - 4. Simplify further by factoring it using the difference of squares:
4x(x - 2)(x + 2) ≥ 0
To determine the values of x that make the inequality true, we need to consider the sign of each factor:
a. 4x ≥ 0: This is true when x ≥ 0.
b. (x - 2) ≥ 0: This is true when x ≥ 2.
c. (x + 2) ≥ 0: This is true when x ≥ -2.
To satisfy the original inequality, all three conditions should be true simultaneously, so the solution is x ≥ 2.
2. 4 ≥ 2x^3 + 8x
Simplify the equation by dividing both sides by 2 to make it easier to work with:
2 ≥ x^3 + 4x
Start by subtracting 2 from both sides:
0 ≥ x^3 + 4x - 2
Again, factor out x from each term:
0 ≥ x(x^2 + 4) - 2
Simplify further by factoring x^2 + 4 using the sum of squares:
0 ≥ x(x + 2i)(x - 2i) - 2
Since x^2 + 4 = (x + 2i)(x - 2i), we have two complex roots: 2i and -2i. However, the original inequality was in terms of real numbers, so we can ignore the complex roots.
To find the values of x that make the inequality true, consider the sign of the remaining factor:
x(x - 2i)(x + 2i) - 2 ≥ 0
Since we only have one factor left, we need to consider two cases based on its sign:
a. x ≥ 0: In this case, the inequality is true for any non-complex solution.
b. x < 0: In this case, the inequality becomes x(x + 2i)(x - 2i) - 2 ≥ 0, which is also true for any non-complex solution.
Thus, the solution to the inequality is all real numbers, x ∈ (-∞, ∞).