Tracy D.

how do you solve 8x+32/x^2-16

it is radicals what if x has no value and results in denominator with a value of 0

and is not undefined Joseph M.

as I said x cannot equal 4 or -4, since both make the original denominator equal to 0.

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09/09/12 Marvin F.

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).

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05/20/13 Thomas G.

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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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Tracy D.

what if variable has no value and results in a denominator of 0

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08/24/12 Joseph M.

the equation is undefined at x = 4, and x = -4

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09/09/12 Joseph M.

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.

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09/09/12 Dale B.

You can't solve unless there is an = sign somewhere. The other answers help you SIMPLIFY the expression
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07/17/19 Alex L.

Factoring this problem to 8(x+4)/((x+4)(x-4)) is the first step. Next you should recognize that you have an (x+4) factor in the numerator and the denominator. This causes there to be a hole at x= - 4. Next you notice that you still have a factor of (x-4) in the denominator that causes a vertical asymptote at x=4. This graph will look exactly like the graph of 8/(x-4) except it will have a hole at x=-4 or at the coordinate (- 4, -0.5) there will be a hole with no value. As x approaches 4 from the negative side, y will approach negative infinity and as x approaches 4 from the positive side y will approach positive infinity. As x approaches positive and negative infinity, y will simply approach zero.
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