Answer:
x = √17 and x = -√17
Step-by-step explanation:
We have the equation:
![\frac{3}{x + 4} - \frac{1}{x + 3} = \frac{x + 9}{(x^2 + 7x + 12)}](https://tex.z-dn.net/?f=%5Cfrac%7B3%7D%7Bx%20%2B%204%7D%20%20-%20%5Cfrac%7B1%7D%7Bx%20%2B%203%7D%20%20%3D%20%5Cfrac%7Bx%20%2B%209%7D%7B%28x%5E2%20%2B%207x%20%2B%2012%29%7D)
To solve this we need to remove the denominators.
Then we can first multiply both sides by (x + 4) to get:
![\frac{3*(x + 4)}{x + 4} - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2A%28x%20%2B%204%29%7D%7Bx%20%2B%204%7D%20%20-%20%5Cfrac%7B%28x%20%2B%204%29%7D%7Bx%20%2B%203%7D%20%20%3D%20%5Cfrac%7B%28x%20%2B%209%29%2A%28x%20%2B%204%29%7D%7B%28x%5E2%20%2B%207x%20%2B%2012%29%7D)
![3 - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}](https://tex.z-dn.net/?f=3%20%20-%20%5Cfrac%7B%28x%20%2B%204%29%7D%7Bx%20%2B%203%7D%20%20%3D%20%5Cfrac%7B%28x%20%2B%209%29%2A%28x%20%2B%204%29%7D%7B%28x%5E2%20%2B%207x%20%2B%2012%29%7D)
Now we can multiply both sides by (x + 3)
![3*(x + 3) - \frac{(x + 4)*(x+3)}{x + 3} = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}](https://tex.z-dn.net/?f=3%2A%28x%20%2B%203%29%20%20-%20%5Cfrac%7B%28x%20%2B%204%29%2A%28x%2B3%29%7D%7Bx%20%2B%203%7D%20%20%3D%20%5Cfrac%7B%28x%20%2B%209%29%2A%28x%20%2B%204%29%2A%28x%2B3%29%7D%7B%28x%5E2%20%2B%207x%20%2B%2012%29%7D)
![3*(x + 3) - (x + 4) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}](https://tex.z-dn.net/?f=3%2A%28x%20%2B%203%29%20%20-%20%28x%20%2B%204%29%20%20%3D%20%5Cfrac%7B%28x%20%2B%209%29%2A%28x%20%2B%204%29%2A%28x%2B3%29%7D%7B%28x%5E2%20%2B%207x%20%2B%2012%29%7D)
![(2*x + 5) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}](https://tex.z-dn.net/?f=%282%2Ax%20%2B%205%29%20%20%3D%20%5Cfrac%7B%28x%20%2B%209%29%2A%28x%20%2B%204%29%2A%28x%2B3%29%7D%7B%28x%5E2%20%2B%207x%20%2B%2012%29%7D)
Now we can multiply both sides by (x^2 + 7*x + 12)
![(2*x + 5)*(x^2 + 7x + 12) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)](https://tex.z-dn.net/?f=%282%2Ax%20%2B%205%29%2A%28x%5E2%20%2B%207x%20%2B%2012%29%20%20%3D%20%5Cfrac%7B%28x%20%2B%209%29%2A%28x%20%2B%204%29%2A%28x%2B3%29%7D%7B%28x%5E2%20%2B%207x%20%2B%2012%29%7D%2A%28x%5E2%20%2B%207x%20%2B%2012%29)
![(2*x + 5)*(x^2 + 7x + 12) = (x + 9)*(x + 4)*(x+3)](https://tex.z-dn.net/?f=%282%2Ax%20%2B%205%29%2A%28x%5E2%20%2B%207x%20%2B%2012%29%20%20%3D%20%28x%20%2B%209%29%2A%28x%20%2B%204%29%2A%28x%2B3%29)
Now we need to solve this:
we will get
![2*x^3 + 19*x^2 + 59*x + 60 = (x^2 + 13*x + 3)*(x + 3)](https://tex.z-dn.net/?f=2%2Ax%5E3%20%2B%2019%2Ax%5E2%20%2B%2059%2Ax%20%2B%2060%20%3D%20%20%28x%5E2%20%2B%2013%2Ax%20%2B%203%29%2A%28x%20%2B%203%29)
![2*x^3 + 19*x^2 + 59*x + 60 = x^3 + 16*x^2 + 42*x + 9](https://tex.z-dn.net/?f=2%2Ax%5E3%20%2B%2019%2Ax%5E2%20%2B%2059%2Ax%20%2B%2060%20%3D%20%20x%5E3%20%2B%2016%2Ax%5E2%20%2B%2042%2Ax%20%2B%209)
Then we get:
![2*x^3 + 19*x^2 + 59*x + 60 - ( x^3 + 16*x^2 + 42*x + 9) = 0](https://tex.z-dn.net/?f=2%2Ax%5E3%20%2B%2019%2Ax%5E2%20%2B%2059%2Ax%20%2B%2060%20-%20%28%20%20x%5E3%20%2B%2016%2Ax%5E2%20%2B%2042%2Ax%20%2B%209%29%20%3D%200)
![x^3 + 3x^2 + 17*x + 51 = 0](https://tex.z-dn.net/?f=x%5E3%20%2B%203x%5E2%20%2B%2017%2Ax%20%2B%2051%20%3D%200)
So now we only need to solve this.
We can see that the constant is 51.
Then one root will be a factor of 51.
The factors of -51 are:
-3 and -17
Let's try -3
p( -3) = (-3)^3 + 3*(-3)^2 + +17*(-3) + 51 = 0
Then x = -3 is one solution of the equation.
But if we look at the original equation, x = -3 will lead to a zero in one denominator, then this solution can be ignored.
This means that we can take a factor (x + 3) out, so we can rewrite our equation as:
![x^3 + 3x^2 + 17*x + 51 = (x + 3)*(x^2 + 17) = 0](https://tex.z-dn.net/?f=x%5E3%20%2B%203x%5E2%20%2B%2017%2Ax%20%2B%2051%20%3D%20%28x%20%2B%203%29%2A%28x%5E2%20%2B%2017%29%20%3D%200)
The other two solutions are when the other term is equal to zero.
Then the other two solutions are given by:
x = ±√17
And neither of these have problems in the denominators, so we can conclude that the solutions are:
x = √17 and x = -√17