Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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?

Answer 7

-2/5 and 4/3

Answer 8

yes

Answer 9

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.

Answer 10

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.

Answer 11

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

Answer 12

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.

Answer 13

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...

Answer 14

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.

Answer 15

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.

Answer 16

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

Answer 17

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

Answer 18

so far this is all we have

Answer 19

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. :)

Answer 20

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.

Answer 21

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

Answer 22

is that the answer ?