Question

Will input question once this opens with the Equation Button was allowed at the beginning. Will input question once this opens with the Equation Button was allowed at the beginning. @Mathematics

Answer 1

Solve each inequality. Express your answer in interval notation. (Please show me the work in this I am still confused on how to get the answers.) Thanks for any assistance.

\[3x^2+8x \le0\]

Answer 2

\[x(3x+8)=0\]
therefore x=0 and x=-8/3. When x<(-8/3) or x>0 \[3x^2+8x>0\] therefore the interval is [-8/3;0]

Answer 3

How do you know which direction the inequality goes?

Answer 4

thought it would be x>-8/3 and x<0

Answer 5

It should be x < -8/3 or x > 0

Answer 6

take any value from interval [-8/3;0] and you will see that 3x^2+8x<=0 and if you take for example 2, 3x^2+8x will be >0

Answer 7

that's not the way it should be derived ..

Answer 8

but still x>0 or x<-8/3 is not truth

Answer 9

I haven't read the whole thing .. I meant that \[ 3x^2+8x>0 \Rightarrow \left.x<-\frac{8}{3}\right\|x>0 \]

Answer 10

then it would be right :)

Answer 11

and for the first thing \[ -\frac{8}{3}\leq x\leq 0 \]

Answer 12

The problem started as
\[3x^2+8x \le0\]

Answer 13

btw why you did it like that .. this is so simple .. just factor the main thing and you will see you answer instantly!

Answer 14

if i did for myself, i'd have factored, but the way i did is better for explanation

Answer 15

How so? :)

Answer 16

I factored it but couldn't figure out which direction the > or < symbol went. I am guessing it has to do with the division switching the direction of the > <.

Answer 17

No .. say for < 0 two one can positive and one have to be negative ..