(x+1/x-2)^2+x+1/x-4-3(2x-4/x-4)^2=0

5 min read Jun 16, 2024

Solving the Equation: (x+1/x-2)^2 + x+1/x-4 - 3(2x-4/x-4)^2 = 0

This equation presents a challenge due to its complex structure. We will break down the steps to solve it:

a. Factor out common terms:

Observe that both the first and third terms have a common factor of (x-4) in the denominator. We can simplify the equation by factoring this out:

(x+1/x-2)^2 + (x+1)/(x-4) - 3(2x-4/(x-4))^2 = 0

(x+1/x-2)^2 + (x+1)/(x-4) - 3(2(x-2)/(x-4))^2 = 0

b. Simplify the expressions:

Now, we can further simplify by expanding the squares:

[(x+1)^2/(x-2)^2] + (x+1)/(x-4) - 3[4(x-2)^2/(x-4)^2] = 0

c. Combine terms with the same denominator:

To combine the terms, we need to find a common denominator. The common denominator for all terms is (x-2)^2(x-4)^2.

[(x+1)^2(x-4)^2 + (x+1)(x-2)^2(x-4) - 12(x-2)^2(x-4)] / [(x-2)^2(x-4)^2] = 0

Since the denominator cannot be zero, we only need to solve the numerator:

(x+1)^2(x-4)^2 + (x+1)(x-2)^2(x-4) - 12(x-2)^2(x-4) = 0

a. Factor out common terms:

Notice that all terms have a common factor of (x-2)^2(x-4). We can factor this out:

(x-2)^2(x-4)[(x+1)^2 + (x+1)(x-2) - 12] = 0

b. Simplify the expression inside the brackets:

Expanding the expressions inside the brackets, we get:

(x-2)^2(x-4)[x^2 + 2x + 1 + x^2 - x - 2 - 12] = 0

(x-2)^2(x-4)[2x^2 + x - 13] = 0

Now we have a simpler equation to solve. We need to find the values of x that make this equation true:

To solve the quadratic equation 2x^2 + x - 13 = 0, we can use the quadratic formula:

x = [-b ± √(b^2 - 4ac)] / 2a

Where a = 2, b = 1, and c = -13.

Solving for x using the quadratic formula, we get two more solutions:

It's important to verify if the solutions we found are valid. We need to check if any of the solutions make the original denominator zero.

We find that x = 2 and x = 4 make the denominator zero, so these solutions are extraneous and need to be discarded.

Therefore, the solutions to the equation (x+1/x-2)^2 + x+1/x-4 - 3(2x-4/x-4)^2 = 0 are: