Answer:
x = √17 and x = -√17
Step-by-step explanation:
We have the equation:
[tex]\frac{3}{x + 4} - \frac{1}{x + 3} = \frac{x + 9}{(x^2 + 7x + 12)}[/tex]
To solve this we need to remove the denominators.
Then we can first multiply both sides by (x + 4) to get:
[tex]\frac{3*(x + 4)}{x + 4} - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}[/tex]
[tex]3 - \frac{(x + 4)}{x + 3} = \frac{(x + 9)*(x + 4)}{(x^2 + 7x + 12)}[/tex]
Now we can multiply both sides by (x + 3)
[tex]3*(x + 3) - \frac{(x + 4)*(x+3)}{x + 3} = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}[/tex]
[tex]3*(x + 3) - (x + 4) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}[/tex]
[tex](2*x + 5) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}[/tex]
Now we can multiply both sides by (x^2 + 7*x + 12)
[tex](2*x + 5)*(x^2 + 7x + 12) = \frac{(x + 9)*(x + 4)*(x+3)}{(x^2 + 7x + 12)}*(x^2 + 7x + 12)[/tex]
[tex](2*x + 5)*(x^2 + 7x + 12) = (x + 9)*(x + 4)*(x+3)[/tex]
Now we need to solve this:
we will get
[tex]2*x^3 + 19*x^2 + 59*x + 60 = (x^2 + 13*x + 3)*(x + 3)[/tex]
[tex]2*x^3 + 19*x^2 + 59*x + 60 = x^3 + 16*x^2 + 42*x + 9[/tex]
Then we get:
[tex]2*x^3 + 19*x^2 + 59*x + 60 - ( x^3 + 16*x^2 + 42*x + 9) = 0[/tex]
[tex]x^3 + 3x^2 + 17*x + 51 = 0[/tex]
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:
[tex]x^3 + 3x^2 + 17*x + 51 = (x + 3)*(x^2 + 17) = 0[/tex]
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
