it is radicals what if x has no value and results in denominator with a value of 0
and is not undefined
it is radicals what if x has no value and results in denominator with a value of 0
and is not undefined
Solve: 8x + 32 / x^{2} - 16
First factor out as much as possible.
The numerator can be factored out as: 8(x + 4)
The demoninator can be factored out as: (x + 4) (x - 4)
Now you have 8 (x + 4) / (x + 4) (x - 4)
The (x + 4) in both numerator and demoninator can be eliminated.
What you have left is 8 / (x - 4)
what if variable has no value and results in a denominator of 0
the equation is undefined at x = 4, and x = -4
just because the x+4 in the denominator can be factored out does not mean it does not make the denominator equal to 0: it does. Any number that makes the denominator equal to 0 makes the function undefined, so x cannot equal that number, so in this case x cannot equal 4 or -4.
Everthing said by the other posters here is correct with one addition just as x cannot equal 4, x cannot equal -4, because the original denominator was (x-4)(x+4), and neither factor can equal 0. x-4 = 0, x=4, x+4 = 0, x=-4. There is a little jump or a circle on the graph when x=-4.
To solve 8x+32/x^{2}-16 factor both the denominator x^{2}-16 and the numerator 8x+32 as follows:
8x+32=8(x+4)
x^{2}-16= (x+4)(x-4)
Therefore, 8x+32/x2-16=8(x+4)/(x+4)(x-4)
Simplify, by dividing both the numerator and the denominator by the common factor,(x+4)
The final answer is 8/x-4. Solution: All real numbers except x=4, which would render the denominator zero and the fraction undefined.
If (8x+32)/ (x^2 -16) is meant, then
factor the numenerator: 8(x+4)
factor the denominator: (x+4)(x-4)
resulting in: 8(x+4)/(x+4)(x-4)
the (x+4) term in numerator and denominator cancel, leaving 8/(x-4).
The solution for x is all real numbers, except x=4.
how do you solve 8x+32/x^2-16
To solve this you must use factoring.
Factor 8x+32 = 8(x+4)
Factor X^2-16 = (x-4)(x+4)
Now Divide = 8(x+4)/(x-4)(x+4)
Simplify = (x+4) cancels out
Answer = 8/(x-4)
There are a few ways SET 8x+32/x^2-16=0
Then multiply the equation by x^2 -> 8x^3+32-16x^2=0
Take the derivative (if you know calculus) -> 24x^2-16x=0 and divide by 16 x^2-2/3 x=0
the derivative set to zero give points of zero slope for the equation not the zeros on the x axis You can plot by hand or use a graphing calculator to find the root which is near -1
Comments
as I said x cannot equal 4 or -4, since both make the original denominator equal to 0.
The question is incomplete. In order to find x, an equation is needed. An equation must have an "=" sign. It would also help if you could clarify if you mean 32/x^2 or 32/x^(2-16).
Marvin makes an excellent point- we need an equation! Since this is an algebra problem and the x-y coordinate system is such a large part of algebra, it would be a valid assumption to equate the given expression to y. Now we need to decide what the form of the expression is, since it isn't clear.
Three possibilites are: 1. 8x + (32/x^2) - 16. (three terms)
2. 8x +32/(x^2 -16). (two terms)
3. (8x+32)/(x^2 - 16). (one term)
Since no parentheses were used in the original expression I think my first possibility is most probable. Therefore, to solve the first possibility I set the expression equal to y. y = 8x + (32/x^2) - 16. Then I graphed the equation to find the solutions for any x or y. We can see of course that x cannot = 0, since that would make the second term undefined. The most noticeable characteristics of the graph are the asymptotes that approach x=0 from the positive and negative directions.
The other two possibilities could be handled in similar fashion.
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