Question

Himpunan penyelesaian pertidaksamaan nilai mutlak |2x-4| > |3x+2|

Answer 1

Answer:

The solution in interval notation is:

$(-6,\frac{2}{5})$.

The solution in inequality notation is:

$-6$.

Step-by-step explanation:

I think you are asking how to solve this for $x$.

Keep in mind $|x|=\sqrt{x^2}$.

$|2x-4|>|3x+2|$

$\sqrt{(2x-4)^2}>\sqrt{(3x+2)^2}$

If $\sqrt{u}>\sqrt{v}$ then $u>v$.

$(2x-4)^2>(3x+2)^2$

Subtract $(2x-4)^2$ on both sides:

$0>(3x+2)^2-(2x-4)^2$

Factor the difference of squares $a^2-b^2=(a-b)(a+b)$:

$0>((3x+2)-(2x-4))((3x+2)+(2x-4))$

Simplify inside the factors:

$0>(x+6)(5x-2)$

$(x+6)(5x-2)$

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 $(x+6)(5x-2)$ 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:

$(-6,\frac{2}{5})$.

The solution in inequality notation is:

$-6$.