inequalities

[smokin92]

3x+4<0

AND

xsquared-3x-12<0

Could someone please walk me through how to do these, I kinda/sorta know how, but i just want to make sure.

Thanks [/tex][/list]

 

[skeeter]

you work the first one (easy), and I'll show you how the second one (little harder) works.

 

[smokin92]

ok. first you solve it as a regular equation right?

like 3x + 4 = 0
-4 -4
3x = -4
-- --
3 3

x= -4
---
3


Then sub it into the inequality...

3*-4/3 +4 < 0
-4 +4 < 0
0 <0

:?

then what would be the answer...
Im thinking theres another step or something,

Sorry if I dont know anything about this lol, I never really got it when the teacher explained it.

 

[skeeter]

you're making this one way too hard when it is extremely simple to do ...
3x + 4 < 0
3x < -4
x < -4/3

now for the quadratic inequality ...
x² - 3x - 12 < 0
factor ...
(x - 4)(x + 3) < 0
at x = 4 and x = -3, the left side of the inequality equals 0 ... that means that every other value of x makes the left side either positive or negative.

the values -3 and 4 break up the x-values into 3 regions ...
check any value in each region to see if it makes the left side positive or negative ...

<---------|--------------|--------->
.............-3................4.............

for x < -3, try x = -4 ... (-4 - 4)(-4 + 3) > 0 , all values in this region make the inequality false
for -3 < x < 4, try x = 0 ... (0 - 4)(0 + 3) < 0 , all values in this region make the inequality true
for x > 4, try x = 5 ... (5 - 4)(5 + 3) > 0 , all values in this region make the inequality false.

so ... the solution set for the quadratic inequality is -3 < x < 4

now, my question is this ...
are you supposed to find the solution set that satisfies both inequalities?
if so, then determine where they overlap.

 

[smokin92]

Nope, there different problems, thank you soo much for the help.