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)
Solve the inequality and express the answer in interval notation:
1.) 3x^2+2x>2x^2-3
2.) 4x^2-x<0
3 answers
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
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
1 answer