1. ## inequality

Solve the inequality below by using algebra. You must show your work (for the test) on this problem to receive full credit. Type in your answer using interval notation (using inf for infinity and u for union).
6x2+3x<4-2x

for an answer i got, (-inf,-4/3)u(1/2,inf).

it says I'm in correct, I'm at a complete loss. I'm almost positive the algebra is right, but I think I'm screwing up writing it out in interval notation.

2. Originally Posted by Twan
Solve the inequality below by using algebra. You must show your work (for the test) on this problem to receive full credit. Type in your answer using interval notation (using inf for infinity and u for union).
6x2+3x<4-2x

for an answer i got, (-inf,-4/3)u(1/2,inf).

it says I'm in correct, I'm at a complete loss. I'm almost positive the algebra is right, but I think I'm screwing up writing it out in interval notation.

Show us what you did.

3. Originally Posted by Drexel28
Show us what you did.

6x^2+5x-4<0

(6x^2+8x)(-3x-4)

2x(3x+4)-1(3x+4)

2x-1<0 3x+4<0
x<1/2 x<-4/3

4. Hello, Twan!

Solve the inequality: . $6x^2 + 3x \:<\:4 - 2x$

We have: . $6x^2 + 5x - 4 \:<\:0$

The graph of: . $y \:=\:6x^2 + 5x - 4$ is an up-opening parabola: $\cup$
. . It is negative between its x-intercepts.

To find the x-intercepts: . $6x^2 + 5x - 4 \:=\:0\quad\Rightarrow\quad (3x+4)(2x-1) \:=\:0 \quad\Rightarrow\quad x\:=\:\text{-}\tfrac{4}{3},\:\tfrac{1}{2}$

Therefore, the interval is: . $\left(\text{-}\frac{4}{3},\:\frac{1}{2}\right)$

5. oh ic, thank you very much, i thought i'd have to do a union. If that problem was altered slightly what should i look for if there's a possible union or just a simple point like (-4/3,1/2)

6. Originally Posted by Twan
oh ic, thank you very much, i thought i'd have to do a union. If that problem was altered slightly what should i look for if there's a possible union or just a simple point like (-4/3,1/2)
This is not a point! This is the interval $\frac{-4}{3}!

7. 6x^2+5x-4<0
=>(x-1/2)(x+4/3)<0
=>either (x-1/2)<0 and (x+4/3)>0
or (x-1/2)>0 and (x+4/3)<0 (not possible)
=> -4/3 <x<1/2
=>x belongs to (-4/3,1/2) which is the required interval.