Respuesta :
Answer:
The solution in interval notation is:
[tex](-6,\frac{2}{5})[/tex].
The solution in inequality notation is:
[tex]-6<x<\frac{2}{5}[/tex].
Step-by-step explanation:
I think you are asking how to solve this for [tex]x[/tex].
Keep in mind [tex]|x|=\sqrt{x^2}[/tex].
[tex]|2x-4|>|3x+2|[/tex]
[tex]\sqrt{(2x-4)^2}>\sqrt{(3x+2)^2}[/tex]
If [tex]\sqrt{u}>\sqrt{v}[/tex] then [tex]u>v[/tex].
[tex](2x-4)^2>(3x+2)^2[/tex]
Subtract [tex](2x-4)^2[/tex] on both sides:
[tex]0>(3x+2)^2-(2x-4)^2[/tex]
Factor the difference of squares [tex]a^2-b^2=(a-b)(a+b)[/tex]:
[tex]0>((3x+2)-(2x-4))((3x+2)+(2x-4))[/tex]
Simplify inside the factors:
[tex]0>(x+6)(5x-2)[/tex]
[tex](x+6)(5x-2)<0[/tex]
The left hand side is a parabola that faces up. I know this because the degree is 2.
The zeros of the the parabola are at x=-6 and x=2/5.
We can solve x+6=0 and 5x-2=0 to reach that conclusion.
x+6=0
Subtract 6 on both sides:
x=-6
5x-2=0
Add 2 on both sides:
5x=2
Divide both sides by 5:
x=2/5
Since the parabola faces us and [tex](x+6)(5x-2)<0[/tex] then we are looking at the interval from x=-6 to x=2/5 as our solution. That part is where the parabola is below the x-axis. We are looking for where it is below since it says the where is the parabola<0.
The solution in interval notation is:
[tex](-6,\frac{2}{5})[/tex].
The solution in inequality notation is:
[tex]-6<x<\frac{2}{5}[/tex].
