#4x(7-3x)(x-2)<0
40 messages · Page 1 of 1 (latest)
it is correct
I just want the Solution for the first picture.
(so I could know what to write next time.
Just like the second picture where you have S=]-∞;3]U[5,5;+∞[
But I want it for the first picture
it would be S =]-∞; 0[ U ]2; 7/3[
Why you don't put +∞ here?
Oh....... Makes more sense.
I'm going to keep this open for a while until I fully understand the logic.
What did you use for the graph?
Ok thanks.
look at it and see if it makes sense
I'm translating in my mind, might take some time. But thanks.
you're welcome
I went to chatgpt and it gave me a wrong answer lol
this is the right answer
alright i helped him out and he gave me this:
To see why the solution is S = ]-∞; 0[ U ]2; 7/3[, we can use the same sign chart method as before:
Critical points: 0, 7/3, 2
Intervals: (-∞, 0), (0, 2), (2, 7/3), (7/3, ∞)
Test points:
x = -1 (in interval 1): 4*(-3(-1)+7)(-1-2) = 24 > 0
x = 1 (in interval 2): 4*(-3(1)+7)(1-2) = -16 < 0
x = 3/2 (in interval 3): 4*(-3(3/2)+7)(3/2-2) = 3 < 0
x = 4 (in interval 4): 4*(-3(4)+7)(4-2) = 40 > 0
Based on these test points, we can see that the expression is negative in intervals 2 and 3, and positive in intervals 1 and 4. Therefore, the solution to the inequality is:
x ∈ ]-∞, 0[ U ]2, 7/3[
I hope this clears up any confusion, and thank you again for bringing the error to my attention.
@thick parrot So my maths professor told me 4x(7+3x)(x-2)<0 and the answer would be S=] 0;2 [ U ] ⁷/3;+∞ [
But if 4x(7+3x)(x-2)>0 then the answer would be S=] -∞;0 [ U ] 2;⁷/3 [
it is everything except that other interval and the zeros