Solve the inequality and express the answer in interval notation:
1.) 3x^2+2x>2x^2-3
2.) 4x^2-x<0
3x^2+2x > 2x^2-3
x^2 + 2x + 3 > 0
since the discriminant is negative (4-12), this has no real roots. That is, it does not cross the x-axis. Since f(0) = 3 > 0, f(x) is always > 0, so the solution is all real numbers.
4x^2 - x < 0
x(4x-1) < 0
Recall what you know about parabolas. Since this one opens upward, f(x) is negative between the roots. The roots are 0 and 1/4, so f(x) < 0 in (0,1/4)
x^2 + 2 x + 3 > 0
upward opening parabola (holds water)
where is the vertex?
complete square
x^2 + 2 x = y - 3
x^2 + 2 x + 1 = y - 2
(x+1)^2 = y - 2
vertex at (-1 , 2)
so y is never below y = 2
x^2 + 2 x + 3 > 2
3x-1÷4+x-2÷3-x-1÷5<1÷6
To solve the inequalities and express the answers in interval notation, follow these steps:
1. Solve the inequality as you would solve an equation, but keep in mind that when you multiply or divide both sides by a negative number, the direction of the inequality sign will reverse.
2. Simplify the expression to eliminate any parentheses, brackets, or combining like terms.
3. If necessary, move all terms to one side of the inequality sign (usually the left side) to get the inequality in the standard form: ax + b < 0 or ax + b > 0.
4. Factor the expression (if possible) and find the critical points by setting each factor equal to zero and solving.
5. Plot the critical points on a number line and pick test points within each interval.
6. Determine the sign of each interval by plugging in the test points and evaluating the inequality.
7. Express the solutions as intervals.
Now, let's solve the given inequalities:
1.) 3x^2 + 2x > 2x^2 - 3
Start by moving all the terms to one side:
3x^2 + 2x - 2x^2 + 3 > 0
Combine like terms:
x^2 + 2x + 3 > 0
This quadratic equation does not factor easily, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
For our equation, a = 1, b = 2, and c = 3. Plugging these values into the quadratic formula:
x = (-2 ± √(2^2 - 4(1)(3))) / (2(1))
x = (-2 ± √(4 - 12)) / 2
x = (-2 ± √(-8)) / 2
x = (-2 ± 2√(-2)) / 2
x = -1 ± √(-2)
Since we have a square root of a negative number, this inequality has no real solutions. Therefore, there is no interval notation to express the answer.
2.) 4x^2 - x < 0
Start by factoring the equation:
x(4x - 1) < 0
Now, we have two factors, x and (4x - 1). Set each factor equal to zero:
x = 0
4x - 1 = 0
Solving these equations, we get:
x = 0
x = 1/4
Plotting these critical points on a number line:
-∞ 0 1/4 ∞
Now, pick a test point from each interval and plug it into the inequality to determine the sign:
For x < 0, let's choose x = -1:
(-1)(4(-1) - 1) < 0
(-1)(-4 - 1) < 0
(-1)(-5) < 0
5 < 0 (This is false, so the interval (-∞, 0) is not part of the solution set.)
For 0 < x < 1/4, let's choose x = 1/8:
(1/8)(4(1/8) - 1) < 0
(1/8)(1 - 1) < 0
(1/8)(0) < 0
0 < 0 (This is false, so the interval (0, 1/4) is not part of the solution set.)
For x > 1/4, let's choose x = 1:
(1)(4(1) - 1) < 0
1(4 - 1) < 0
1(3) < 0
3 < 0 (This is false, so the interval (1/4, ∞) is not part of the solution set.)
Since none of the intervals satisfy the inequality, the solution set is empty, denoted as ∅ or {}.