Answer:
(-infinity, -1/2] and [3, infinity] Let e know if anything didn't make sense.
Step-by-step explanation:
A good way to think about it is a graph is either at 0, greater than 0 or less than 0. If you can find all of the zeroes then you can test each zero by checking points before and after. the graph will either cross and change signs, or stay the same.
So we want to find the zeroes.
Much like the factored form of a quadratic (x - a)(x - b) this inequality has two expressions being multiplied together, so if you can find when they are 0 you will have all the zeros. We shoudl also check when they do not exist just to be safe.
(x^2 - 3x) itself is a quadratic. so lets find its zeroes.
x^2 - 3x
x(x - 3)
So the zeroes are 0 and 3. lets find the zeroes of the other expression before checking signs.
sqrt(2x^2-3x-2) will be at 0 when the quadratic inside is at 0, and will not exist if the quadrati is negative, so let's look for both. first the 0s. I am going to complete the square
2x^2 - 3x - 2 = 0
2(x^2 - (3/2)x - 1) = 0
2[(x - 3/4)^2 - 25/16] = 0
2(x - 3/4)^2 - 25/8 = 0
2(x - 3/4)^2 = 25/8
(x - 3/4)^2 = 25/16
x - 3/4 = +/- 5/4
x = +/-5/4 + 3/4
x = -2/4 or 8/4
x = -1/2 or 2
Let me know if you couldn't follow that. Anyway, using those zeroes we can tell 2x^2 - 3x - 2 is negative between the two zeroes, since it is a parabola that opens upwards and crosses the x axis at those zeroes. this means sqrt(2x^2 - 3x - 2) does not exist from x=-1/2 and x=2. This means for the other zero above at x=0, the iequality will not exist. So now we have 3 points to look at.
-1/2 for its first 0 and where it then stops existing up to 2, which is also a 0. Finally 3 is the last 0.
at x=-1/2 all values before it are positive
x=2 has values immediately after it negative
x=3 then it switches so all values after that are positive since it is the last 0
Now we know that (x^2 - 3x) * sqrt(2x^2 - 3x - 2)is positive or equal to 0 along the intervals, (-infinity, -1/2] and [3, infinity]
