SOLUTION: 5z/z-2 - 10/z+2 = 40/z^2 -4

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Question 263671: 5z/z-2 - 10/z+2 = 40/z^2 -4
Found 2 solutions by JBarnum, jsmallt9:
Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
5z/z-2 - 10/z+2 = 40/z^2 -4
remeber to multiply every thing by a denomenator that will eleminate x from the denominator. this problem requires us to multiply by ((z-2)(z+2))
5z/z-2((z-2)(z+2)) - 10/z+2((z-2)(z+2)) = 40/z^2 -4((z-2)(z+2))
5z(z+2)-10(z-2)=40 subtract 40 and simplify multiplication
5z^2+10z-10z+20-40=0
5z^2-20=0 divide by 5
z^2-4=0
(z-2)(z+2)=0
Z=(-2,2)
Heres the pluggin solver
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 5x%5E2%2B0x%2B-20+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%280%29%5E2-4%2A5%2A-20=400.

Discriminant d=400 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28-0%2B-sqrt%28+400+%29%29%2F2%5Ca.

x%5B1%5D+=+%28-%280%29%2Bsqrt%28+400+%29%29%2F2%5C5+=+2
x%5B2%5D+=+%28-%280%29-sqrt%28+400+%29%29%2F2%5C5+=+-2

Quadratic expression 5x%5E2%2B0x%2B-20 can be factored:
5x%5E2%2B0x%2B-20+=+5%28x-2%29%2A%28x--2%29
Again, the answer is: 2, -2. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+5%2Ax%5E2%2B0%2Ax%2B-20+%29

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
A solution provided by another tutor is correct as far as it goes. But it is missing a critical last step!

The solution provided multiplies both sides of the equation by (z+2)(z-2). Multiplying both sides of an equation by an expression that might end up being zero (and (z+2)(z-2) might be zero depending on the value fo z) may introduce what are called extraneous solutions. Extraneous solutions are solutions that work in the equation after this multiplication but do not work in the original equation! So we must check our solution(s)!

The other tutor's solutions are 2 and -2. Let's check them:
5z%2F%28z-2%29+-+10%2F%28z%2B2%29+=+40%2F%28z%5E2+-4%29
Checking z = 2:
5%282%29%2F%28%282%29-2%29+-+10%2F%28%282%29%2B2%29+=+40%2F%28%282%29%5E2+-4%29
Simplifying we get:
10%2F0+-+10%2F4+=+40%2F0
As we can see, we get some zero denominators!! So z = 2 is an extraneous solution and must be rejected. (If only one denominator was zero we would still have to reject this solution.)

Checking z = -2:
5%28-2%29%2F%28%28-2%29-2%29+-+10%2F%28%28-2%29%2B2%29+=+40%2F%28%28-2%29%5E2+-4%29
Simplifying we get:
%28-10%29%2F%28-4%29+-+10%2F0+=+40%2F0
Again we get some zero denominators!! So z = -2 is also an extraneous solution and must be rejected. (If only one denominator was zero we would still have to reject this solution.)

We have rejected both solutions! This means that your original equation has no solution!