Help with a rational expression equation?

Question

x / (x-2) - 2 / (x^2 +2x -8) = 5 / (x+4)

Answer 1

x / (x - 2) - 2 / (x + 4) (x - 2) = 5 / (x + 4)
x (x + 4) - 2 = 5 (x - 2)
x² + 4x - 2 = 5x - 10
x² - x + 8 = 0
x = [1 ± √(1 - 32)] / 2
x = [ 1 ± i √31 ] / 2

Answer 2

A not-so-easy one!
x/(x-2)-2/(x^2+2x-8)=5/(x+4)
x/(x-2)-2/(x^2+2x-8)-5/(x+4)=0
x/(x-2)-2/(x-2)(x+4)-5/(x+4)=0
The lowest common factor is (x-2)(x+4)

{x(x+4)-2 -5(x-2)} / (x-2)(x+4) =0
Multiply both sides by (x-2)(x+4),
x(x+4)-2-5(x-2)=0
x^2+4x-2-5x+10=0
x^2-x+8=0 Alas! No factors easily seen.
Use quadratic formula.

x={-b+/-rt(b^2-4ac)} / 2a
a=1, b=-1, c=8
x={-(-1) +/- rt(-1)^2-4(1)(8)} / 2
x=1+/- rt(1-32)}/2
x={1+/- rt(-31)}/2
I'm going to stop right here. Your equation has NO solutions in the Real number system, because I can't take the square root of a negative number.
There is a solution if you use Complex Numbers, but you're not into that yet, are you? That's where you use a letter i, i being = sq.rt (-1)

Going back a few steps, I should have examined my
quadratic equation more closely once I had
x^2-x+8=0.
In the quadratic formula, the term (b^2-4ac)
plays a key role.
It is given the name "DISCRIMINANT", because it determines the nature of the quadratic's solution.
If b^2-4ac is +, you will get two real, unequal roots.
If b^2-4ac is 0, you get two equal roots
If b^2-4ac is -, there are no real roots.
All that I had to do to know where I was heading was to use the a,b, and c values to evaluate the Discriminant. Had I done this, I would have quickly seen b^2-4ac=(-1)^2-4(1)(8), =1-32, =-31. It's negative, therefore no solution possible.
No need to bleed to death slowly, struggling with all the terms in the quadratic formula.

So, my recommendation to you is this;
If you have to use the quadratic formula, calculate
b^2-4ac first. You're going to need it anyway.
If it is positive or zero, carry on: you can take the square root of a + no., or 0. If b^2-4ac is a negative
number, you can report "no solution", because you won't be able to take the square root.
Here endeth the lesson!

Answer 3

x / (x-2) - 2 / (x^2 +2x -8) = 5 / (x+4)

First, factor the trinomial

x / (x-2) - 2 / [(x +4)(x - 2)] = 5 / (x+4)

Now multiply both sides by the LCD (x - 2)(x + 4) to clear the denominators

x(x + 4) - 2 = 5(x - 2)
x^2 + 4x - 2 = 5x - 10
x^2 - x + 8 = 0

Unfortunately, this trinomial doesn't factor easily - y0u will have to use the quadratic formula

x = [-b +- sqrt(b^2 - 4ac)]/2a

Answer 4

16

Answer 5

you do it with the through finding the common component: (x^2*y^2)+(-2x^3*y) --------------------------- take x^2 * y as uncomplicated component 4x^2*y x^2 *y (y)+(-2x) ------------------------ now devide x^2 * y with the only interior the denominator 4x^2*y y - 2x --------------------------- take denominator on my very own on the factor 4 one million ----- ( y - 2x ) right here you bypass 4