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(FIRST PART ALREADY ANSWERED AND NEW PART ADDED) Why do you find the domain of both the numerator and denominator in a rational equation? When do you do it?

For example, if you had:
 
x-3
------
5x+3
 
you'd only find the domain and use the denominator. Why wouldn't you also do the numerator? 
 
And then you'd use both the  numerator and the denominator in this:
 
(x^4-9x^2)^1/2
------------------------
x^2-4
 
WHY DO YOU USE BOTH OF THEM THERE? WHY NOT ON THE FIRST ONE?
 
~NEW QUESTION~
 
So if it was linear, like the first example, then you'd only have to use the denominator to find the domain because, mainly, your only concern is that the denominator may be 0, which then would make the value not real. But for the second example, you have to use both the numerator and the denominator because they contain a root, and roots (if they're even) could make the situation imaginary thus not real. But what if the root in the denominator was a cube root?

For example

cube root (6x+3)
-----------------------
x^2-4

Would you need to use the numerator then?
 
 
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1 Answer

Hi M, as you know the domain is a set of x values involved in a graph.  x -3 and 5x-3, alone, are linear and any number you put in for x, you will get a y which means the graph exist at that point.  Any equation that is linear(y=mx + b) or quadratic (y=x2), cubic,...,with no denominator is the set of all x values because any value you put in the equation will have a y-value.  However when (x-3)/(5x-3), there is a value of x where it would make the denominator 0, which would make the graph undefined at that point.  This is why, in finding the domain, you set 5x-3 not= 0.  There will be a vertical asymptote at that point that the graph will get close to, but never touch. If you have a radical such as the sqrt(x-4) or (x-4)^1/2, square roots cannot be less than 0 or negative, if they are, they are not real numbers.  To solve sqrt(x-4), simply set x-4 greater than or equal to 0.  Another scenario to look out for is square roots in denominator-you might be tempted to set the square root to greater than or equal to 0, but you cannot have a 0 in the denominator so you would set the square root to greater than 0.
 
In summary, denominators can NEVER equal zero and square roots, with real numbers, cannot be less than zero.
 
Best of luck,
Shannon

Comments

So if it was linear, like the first example, then you'd only have to use the denominator to find the domain because, mainly, your only concern is that the denominator may be 0, which then would make the value not real. But for the second example, you have to use both the numerator and the denominator because they contain a root, and roots (if they're even) could make the situation imaginary thus not real. But what if the root in the denominator was a cube root? 
 
For example
 
cube root (6x+3)
-----------------------
x^2-4
 
Would you need to use the numerator then? 
 
Thank you very much. You've already been a great help!
Yes.  So if it was linear, like the first example, then you'd only have to use
the denominator to find the domain because, mainly, your only concern
is that the denominator may be 0, which then would make the value not
real. Yes.  But for the second example, you have
to use both the numerator and the denominator because they contain a
root, and roots (if they're even) could make the situation imaginary
thus not real. But what if the root in the denominator was a cube root? Cube roots or any odd root can be any x value, so no need to use the numerator, then.  If you have any other questions, just look at a graph, you will be able to see where the graph will not be continuous just focusing on the end points.  Best of luck, Shannon
Thanks so much! I totally understand, and now I think I'll do great on my quiz!

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