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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

 

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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6 Answers

Solve: 8x + 32 / x2 - 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)

 

                                                 

 

Comments

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.  

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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/x2-16 factor both the denominator x2-16 and the numerator 8x+32 as follows:

8x+32=8(x+4)

x2-16= (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