Algebraic Equation Solving: Find the Value of a

We are given the equation:

We can rewrite the equation as:

Let’s rearrange the terms on both sides to group them in pairs:

Expand the products within the brackets:

Let and . Then the equation becomes:

Expand both sides:

Subtract 120 from both sides:

Rearrange the terms:

Factor the difference of squares and the common factor:

Factor out the common term :

This equation holds if either or .

Case 1:
Substitute back and :

We need to check if is a valid solution. The original equation has terms in the denominator . If , the denominator is . The numerator is . So, , which is true. Thus, is a solution.

Case 2:
Substitute back and :

These are complex solutions. The problem does not specify whether we are looking for real or complex solutions. If we are looking for real solutions, then there are no solutions from this case. If we are looking for complex solutions, then and are solutions.

Let’s check if the denominator is zero for the complex solutions. The denominator is . If , then the factors are , , , . None of these are zero. Similarly for .

If the problem is intended for real numbers, then the only solution is . If the problem is intended for complex numbers, then there are three solutions: , , and .

Given the context of typical math problems presented in this format, it is likely that only real solutions are expected.

The final answer is .