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Can someone help me with this? Precalc solving inequalities with one variable? The 3x is throwing me off
 one year ago
 one year ago
Can someone help me with this? Precalc solving inequalities with one variable? The 3x is throwing me off
 one year ago
 one year ago

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iluvvyyhuBest ResponseYou've already chosen the best response.1
\[f(x)=(x7)(3x+1)(x+4)\]
 one year ago

Brittni0605Best ResponseYou've already chosen the best response.0
do you need to simplify?
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
You actually have 2 unknowns here. Is there more to this question?
 one year ago

iluvvyyhuBest ResponseYou've already chosen the best response.1
No, you use a sign chart. @Brittni0605 And yes sorry it says *In the exercises 16 determine the x values that cause the polynomial functions to be (a)zero (b) positive and (c)negative @ChmE
 one year ago

Brittni0605Best ResponseYou've already chosen the best response.0
I don't see a chart, ill let @ChmE help you.
 one year ago

iluvvyyhuBest ResponseYou've already chosen the best response.1
Oh no its a sign chart that you make to solve it with lol but okay(:
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
so we set it equal to zero, then less than for negative, then greater than for positive. Ok. Let's start with setting it equal to zero. Do not distribute, leave the equation as is. Go to each ( ) and think what will x need to be to make this zero. Only one needs to be zero for the whole thing to be zero. There should be 3 answers. I'll give you a sec. Let me know if you didn't understand what I just said
 one year ago

iluvvyyhuBest ResponseYou've already chosen the best response.1
I got 7, 1/3 and 4 ....but the 1/3 doesnt look right to me:/
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
Thats correct! So that solves our zero.
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
0<x7 7<x 0<3x+1 1<3x 1/3<x 4<x
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
But here is where it is tricky. Say we use the number 0 which satisfies the first but not the last two. Lets see if we get a positive number. 0<(07)(3(0)+1)(0+4) 0<7*1*4 0<28 no. So the answer for 0<.... is x<4 because that is the lowest. ***I missed typed above this is for the positive****
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
Now what do you think the answer will be for the negative?
 one year ago

iluvvyyhuBest ResponseYou've already chosen the best response.1
ummm im a little confused now :(
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
I don't know how to make a sign chart. I don't ever remember having to do that.
 one year ago

iluvvyyhuBest ResponseYou've already chosen the best response.1
Umm well im not sure i think i kinda get it a little better? Im sorry this is taking long btw, just my teacher never helped me lol
 one year ago

iluvvyyhuBest ResponseYou've already chosen the best response.1
Oh okay I got it now(:
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
0>7*1*4 negative. OMG. Now I need to rethink some things. Im so sorry. Lets try to make a sign chart.
 one year ago

iluvvyyhuBest ResponseYou've already chosen the best response.1
lol okaydw:1352178275962:dw would i put the 1/3 on it though? i dont know lol Its bothering me since its a fraction lol
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
so when it is negative is x<4, 1/3<x<7 so when it is positive it is 4<x<1/3, x>7
 one year ago

iluvvyyhuBest ResponseYou've already chosen the best response.1
okay thats easier for me to understand(:
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
How to solve problems like this. Set equal to zero. Solve each term for the zeros. Put those points on a number line. Pick easy numbers between those points. Test to see if it is positive or negative. Follow the number line like a map to get the solutions for neg and pos.
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
Sorry for leading you estray but sometimes working through a problem and finding mistakes helps us understand a little more about the correct way in terms of what is not right. I think a sign graph is very helpful, I'd continue to use that. Do you think you could do a problem like this by yourself now? Thanks for refreshing me as well. It took a while but we got it !!
 one year ago

iluvvyyhuBest ResponseYou've already chosen the best response.1
Umm yes i think so(: ill practice a little more before i goto sleep lol but thanks a bunch <3
 one year ago

ChmEBest ResponseYou've already chosen the best response.1
I'm in differential equations right now which is Calc 4, and I almost got this question wrong, humbling. Good luck and check your work if you have time, it's very beneficial.
 one year ago

iluvvyyhuBest ResponseYou've already chosen the best response.1
Oh wow but alright(: thanks again <3
 one year ago
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