Finding all real solutions?

Question

Thank you to all of those that answered my previous questions. I really appreciate everyone's help.

I have just one more problem that I have been having trouble with.


x / x^2 -4 + 1/ x+2 = 3

Answer 1

From the equation a^2 - b^2 = (a+b)(a-b)
We can form x^2 -2^2 = (x+2)(x-2)
x/(x+2)(x-2) + 1/(x+2) = 3
Make them all common denominators so solve for x.
A common factor is already (x+2) so we try to get (x+2)(x-2) for every variable in the equation. So we multiply 1/(x+2) by (x-2)/(x-2) = 1 so it doesn't change it. We also multiple 3 by (x+2)(x-2)/(x+2)(x-2) to get a equal demoninator. So now we get:
x/(x+2)(x-2) + (x-2)/(x+2)(x-2) = 3(x+2)(x-2)/(x+2)(x-2)
The denominators are all equal now. So we can just remove them to get an equation.
x+x-2= 3x^2 -12
3x^2 -2x -8 = 0.
When we solve for x by quadratic methods.
(3x +4)(x -2) = 0
So since one of the bracketed values can be 0.
x = -4/3 or 2.

Answer 2

I'm guessing that you have (x/(x²-4)) + (1/(x + 2)) = 3 so just multiply the 2'nd term on the left by
(x-2)/(x-2)
to get
(x/(x²-4)) + ((x-2)/(x²-4)) = 3
Then
x+(x-2) = 3x²-12
2x-2 = 3x²-12
3x²-2x-10 = 0 which is in the 'standard form' ax²+bx+c=0
This isn't going to factor nicely, so use the quadratic formula
x = (-b ± √(b²-4ac))/2a
to get your roots.

HTH

Doug

Answer 3

if it is:
x/(x^2-4) + 1/(x+2) = 3
x + (x-2) = 3(x^2-4)
3x^2 -2x -10 = 0

by quadratic formula:

x = [2 ± √(4 + 120)] / 6
x = [2 ± 2√31] / 6 = 1 ± √31.

Answer 4

x / x^2 -4 + 1/ x+2 = 3

x/x*x-4+1/x+2=3
(x)-4+(1/x)+2=3
x+1/x=5
x^2+1=5
x=2