Let f(x) = x/(x+2) + 3/(x-4)

To find the roots, we will rewrite as one fraction.

==> f(x)= [ x(x-4) + 3(x+2) ] / (x+2)(x-4)

==> f(x) = ( x^2 - 4x + 3x + 6) / (x+2)(x-4)

==> f(x) = (x^2 -x + 6) / (x+2)(x-4)

Now the roots if f(x) are the roots of the numerator.

=> x^2 -x + 6 = 0

==> x1= ( 1 + sqrt(1 - 24) /2 = (1/2) + sqrt23 / 2

==> x2= (1/2) - sqrt23*i /2

Then the roots are:

**x = { (1/2) + (sqrt23 /2) *i and (1/2) - (sqrt23 /2)*i }**

What you have given is an expression. Equating it to 0, the roots have been calculated.

x / (x + 2) + 3 / (x - 4) = 0

=> x( x - 4) + 3(x + 2) = 0

=> x^2 - 4x + 6 + 3x = 0

=> x^2 - x + 6 = 0

The roots of this equation are

[ 1 + sqrt (1^2 - 24) ] /2

=> 1/2 + i*sqrt 23 / 2

and 1/2 - i*sqrt 23 / 2

The required roots of the equation are:

**1/2 + (i*sqrt 23) / 2 and 1/2 - (i*sqrt 23) / 2**

Question: What are the roots of the equation x / (x + 2) + 3 / (x -4) = 0.(edited please).

A:

We multiply both sides by (x+2)(x-4):

x(x-4) + 3(x+2) = 0.

x^2-4x+3x +6 = 0.

x^2-x+6 = 0.

We know that ax^2+bx+c = 0 has roots:

x1 = {-b+(b^2-4ac)^(1/2)}/2a or x2 = {-b-(b^2-4ac)^(1/2)}/2a.

So x^2-x+3 has roots:

x1 = {-(-1)+(1-4*1*6)^(1/2)}/2 = {1+(-23)^(1/2)}/2, or

x2 = {1-(-23)^(1/2)}/2.

**So ****the roots are ****x1 = {1+(-23)^(1/2)}/2 and x2 = {1-(-23)^(1/2)}/2.**