Question 143606: Hello, I am completely stumped on the following 2 questions. Any help would be appeciated.
Question #1 is -6/x - 3/x^2 = 2
Question #2 is - 6/x + 6/x+9 = 1
It says I can solve by any method, but I cannot figure how to solve these with either the quadratic equation or by completing the square. If you can point me in the right direction. Thanks.
Answer by nabla(475)   (Show Source):
You can put this solution on YOUR website! 1. If you are unsure of anything I did, I will be happy to iterate the steps I took. E-mail me at enabla@gmail.com
-6/x - 3/x^2 = 2
-6-3/x=2x
-3/x=2x+6
-3=2x^2+6x
0=2x^2+6x+3
Now use the quadratic formulae:
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| Solved by pluggable solver: SOLVE quadratic equation with variable |
Quadratic equation  (in our case  ) has the following solutons:
For these solutions to exist, the discriminant  should not be a negative number.
First, we need to compute the discriminant :  .
Discriminant d=12 is greater than zero. That means that there are two solutions:  .
Quadratic expression  can be factored:
Again, the answer is: -0.633974596215561, -2.36602540378444.
Here's your graph:
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2. Notice the parentheses I put and make sure that it is correct.
- 6/x + 6/(x+9) = 1
-6+6x/(x+9)=x
6x/(x+9)=x+6
6x=(x+6)(x+9)
6x=x^2+15x+54
0=x^2+9x+54
Again, use the quadratic formulae.
| Solved by pluggable solver: SOLVE quadratic equation with variable |
Quadratic equation (in our case  ) has the following solutons:
For these solutions to exist, the discriminant should not be a negative number.
First, we need to compute the discriminant :  .
The discriminant -135 is less than zero. That means that there are no solutions among real numbers.
If you are a student of advanced school algebra and are aware about imaginary numbers, read on.
In the field of imaginary numbers, the square root of -135 is + or -  .
The solution is 
Here's your graph:
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