OpenStudy (anonymous):

guys how can I show (5x+2)(3x-4)≤0 on a number line?

4 years ago
OpenStudy (anonymous):

can you guys draw it on paint or something and upload the picture to internet and then give me link?

4 years ago
OpenStudy (anonymous):

Start by distributing, then you'll get a function in the ax^2 + bx + c form.

4 years ago
OpenStudy (anonymous):

15x^2-14x-8... then what?

4 years ago
OpenStudy (anonymous):

Wait, is this a 1-d line they are talking about?

4 years ago
OpenStudy (anonymous):

http://www.mathematic.ws/wp-content/uploads/2009/04/number-line.png that thing

4 years ago
OpenStudy (anonymous):

Ok, so we are just talking about a number line. For this one, we want to not distribute first. We should look at the two parts separately as well. When does (5x+2)=0 and when does (3x-4)=0?

4 years ago
OpenStudy (anonymous):

-2/5 and 4/3

4 years ago
OpenStudy (texaschic101):

yes

4 years ago
OpenStudy (anonymous):

good, those are our 2 main x values that we will be looking at. at each of those numbers, we get a 0 in the parenthesis, which causes the whole equation to go to 0.

4 years ago
OpenStudy (anonymous):

now that we know when the equation is = to 0, we need to test where it is not. The easiest way to do this, is plug in values. Try plugging in a value, less than -2/5, one between -2/5 and 4/3, and one greater than 4/3.

4 years ago
OpenStudy (anonymous):

yes, so how do we show this situation on the number line

4 years ago
OpenStudy (anonymous):

when you plug in these values, it will be either greater than 0, or less than 0. Compare these answers to your original equation to determine where on the graph it is always less than 0.

4 years ago
OpenStudy (anonymous):

the exact question is like that '' solve by considering all cases. Show each solution on a number line.'' and 15x^2-14x-8≤0 is the question...

4 years ago
OpenStudy (debbieg):

Right, you need to find the solution set - @Mathis1 is walking you through how to do that - then just graph the interval(s) of the solution set on the number line.

4 years ago
OpenStudy (anonymous):

That's fine. because we have found all the places where the equation is = to 0, (-2/5 and 4/3) all the x values in between those values have to be greater than 0 or less than 0.

4 years ago
OpenStudy (anonymous):

by plugging in values, we can see which intervals are greater than 0, and which are less than 0.

4 years ago
OpenStudy (anonymous):

can you guys put it on the number line with paint (the program) etc?

4 years ago
OpenStudy (anonymous):

so far this is all we have |dw:1380391737046:dw|

4 years ago
OpenStudy (debbieg):

Here's a helpful hint, though, with quadratics. Since you know that there are two roots, you know that the pattern of the graph is EITHER: + 0 - 0 + OR - 0 + 0 - and which one it is depends simply on whether the parabola opens up, or opens down. :) Nothing wrong with substituting values to find the signs, but recognizing the behavior of a quadratic function can save some of that effort. :)

4 years ago
OpenStudy (anonymous):

So we know where the endpoints of the graph are going to be since we solved for 0. Those two numbers (-2/5 ans 4/3) are going to be very important. Like Debbie is saying, we need to find out what the values in between those values are, and the values on either side of them.

4 years ago
OpenStudy (anonymous):

So try plugging in values less than -2/5, a value between -2/5 and 4/3, and one greater than 4/3. Pay close attention to their sign (either negative or positive). A helpful hint, you can choose any number, so choose one that is easy to use. I would use -1, 0, and 2

4 years ago
OpenStudy (anonymous):

|dw:1380391879240:dw| is that the answer ?

4 years ago