Solving inequalities algebraically, when root is 0

[supernova1203]

Homework Statement

Solve each inequality without graphing the corresponding function.

State the solution algebraically and graph on a number line:

x/x²-9≤0

so i factor out the denominator and get (x+3)(x-3) the root here is zero, but for some reason in the chart (for rational/reciprocal functions) they seem to treat the vertical asymptotes (x=-3, x=3) as root as well... so really instead of having 1 root, of 0, you now have what look like 3 roots, at x=0, x=-3, x=+3

Now I've only learned how to solve inequalities algebraically this morning from this very helpful youtube video:http://www.youtube.com/watch?v=a9dzsIxcI-o&list=FLU9AMIFm9OGP9S3RGgO_hjw&index=1

and according to this 'method' which i like VERY much...i keep getting the wrong answer/sum for the positives/negative intervals if the zero is equal to 0the book already shows me what the solution is i just don't know how to deal with these roots of 0 in situations like these, can anyone help?
anything would be greatly appreciated, thanks :)

 

[altamashghazi]

for zero solution mark zero on the number line and proceed

 

[supernova1203]

for zero solution mark zero on the number line and proceed

oh hey...i think your right...even if just move in the general direction it will tell me weather or not its positive or negative...

and then i can multiply all the postives/negatives and determine weather the graph in that interval is above 0 or below 0




thanks

 

[SammyS]

Homework Statement

Solve each inequality without graphing the corresponding function.

State the solution algebraically and graph on a number line:

x/x²-9≤0

You have written your algebraic expression incorrectly.

x/x²-9 is technically equivalent to \displaystyle \ \frac{x}{x^2}-9\ which is \displaystyle \ \frac{1}{x}-9\ .

You need to use parentheses and write x/(x²-9) which is \displaystyle \ \frac{x}{x^2-9}\ .

so i factor out the denominator and get (x+3)(x-3) the root here is zero, but for some reason in the chart (for rational/reciprocal functions) they seem to treat the vertical asymptotes (x=-3, x=3) as root as well... so really instead of having 1 root, of 0, you now have what look like 3 roots, at x=0, x=-3, x=+3

Now I've only learned how to solve inequalities algebraically this morning from this very helpful youtube video:

http://www.youtube.com/watch?v=a9dzsIxcI-o&list=FLU9AMIFm9OGP9S3RGgO_hjw&index=1

and according to this 'method' which i like VERY much...i keep getting the wrong answer/sum for the positives/negative intervals if the zero is equal to 0

the book already shows me what the solution is i just don't know how to deal with these roots of 0 in situations like these, can anyone help?

anything would be greatly appreciated, thanks :)

For a rational expression, the critical points are the roots of the numerator together with the roots of the denominator.

This is because the sign of a rational expression will change if either the numerator or denominator changes sign.