+0  
 
0
2
3
avatar+1279 

Find all values of x such that \frac{x}{x - 5} = \frac{4}{2x - 4} + \frac{2}{x + 3}.

 Jan 1, 2024
 #1
avatar+10 
+1

Hey,

I got the equation: x^3-3x^2+12x+10=0.

The problem is that this equation is not factorable with rational numbers and I don't think they want you to use the cubic formula.

Is there something wrong with the problem?

 Jan 2, 2024
 #2
avatar+126665 
+1

\( \frac{x}{x - 5} = \frac{4}{2x - 4} + \frac{2}{x + 3} \)

 

We can simplify to :

 

  x ( 2x - 4) (x + 3)  - 4 (x  - 5)(x + 3)  - 2(x - 5) (2x - 4)  =   0     simplify more

 

2x^3 - 6x^2 + 24x + 20  = 0

 

x^3 - 3x^2 + 12x + 10 =  0

 

By graph, the real solution is ≈  -.6879

 

cool cool cool

 Jan 2, 2024

3 Online Users

avatar