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

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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.

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

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## 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.