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The equation \(\frac{x}{x+1} + \frac{x}{x+2} = kx\) has exactly two different complex solutions. Find all possible complex values for \(k\).

Basically I expanded it and stuff and then made a quadratic and used the discriminant to get \((3k-2)^2-4k(2k-3) <0\). So \(k^2<-4\). But the problem wants all complex solutions so some numbers in the range apparently don't work. 

 Mar 29, 2024
 #1
avatar+474 
0

To solve the equation, we first try to get rid of the denominators.

 

Multiplying both sides of the equation by (x+1)(x+2), we get [x^2 + 2x + x^2 + x = kx(x + 1)(x + 2).]

 

Expanding the right side, we get [2x^2 + 3x = kx^3 + kx^2 + 2kx.]

 

Moving all the terms to one side, we get [0 = kx^3 + (k - 1)x^2 + (2k - 3)x.]

 

Since we are looking for complex solutions, we can treat x as a complex number.

 

For the equation to have exactly two complex solutions, the discriminant of the quadratic part must be nonzero.

 

That is, [(k - 1)^2 - 4 \cdot k \cdot (2k - 3) > 0.]

 

Expanding, we get [k^2 - 2k + 1 - 8k^2 + 24k - 12 > 0.]

 

This simplifies to [-7k^2 + 22k - 11 > 0.]

 

We can factor this as [(k - 1)(7k - 11) > 0.]

 

Thus, k must be in one of the intervals (−∞,1/7)∪(11/7,∞).

 

Now, let's check that for these values of k, the equation has exactly two complex solutions.

 

For k=0, the equation becomes x+1x​+x+2x​=0, which has the complex solutions x=−1 and x=−2

 

For k=2, the equation becomes x+1x​+x+2x​=2x, which has the complex solutions x=0 and x=−23​.

 

Thus, the ony possible k is 11/7.

 Mar 29, 2024
 #2
avatar+128794 
+1

x( x + 2)  + x ( x+ 1) =  kx (x + 1) (x + 2)

 

x^2 + 2x  + x^2 + x  =  kx ( x^2 + 3x + 2)

 

2x^2 + 3x  = kx^3 + 3kx^2 +  2kx

 

kx^3 + (3k-2)x^2 + (2k - 3)x  =  0

 

x [ kx^2  + (3k -2)x + (2k -3)  = 0

 

(3k - 2)^2  - 4 [k *(2k -3)]  < 0

 

9k^2 - 12k + 4 - 8k^2 + 12k  < 0

 

k^2 + 4  <  0

 

k^2  <   -4

 

k < 2i      and  k > -2i

 

k = (-2i, 2i) 

 

cool cool cool

 Mar 29, 2024

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