Solving $a^2 X + (a-1) = (a+1)x$ A Step-by-Step Guide

Jul 27, 2025 by ADMIN 54 views

Solving the Equation $a^2 x + (a-1) = (a+1)x$ : A Comprehensive Guide

Hey guys! Let's dive into solving this intriguing algebraic equation: $a^2 x + (a-1) = (a+1)x$ . This equation, while seemingly simple, opens up a fascinating discussion in mathematics, touching upon various concepts like linear equations, parameter analysis, and solution classification. In this comprehensive guide, we'll break down the steps to solve this equation, explore different scenarios based on the value of 'a', and understand the nuances of the solution set. Get ready to flex those mathematical muscles!

Understanding the Basics

Before we jump into solving for x, let's first get a grip on the key elements of the equation. We have $a^2 x + (a-1) = (a+1)x$ , where x is the variable we want to find, and a is a parameter. A parameter, in this context, is a constant value that can affect the solution of the equation. Think of a as a dial that, when turned, changes the landscape of the solution. Understanding the role of a is crucial because the solution for x will depend on its value. Our main goal here is to isolate x on one side of the equation. We want to manipulate the equation using algebraic operations to get x by itself. This typically involves moving terms around, factoring, and dividing. But we need to be careful! Certain values of a can lead to special cases, like division by zero, which we need to watch out for. The beauty of this equation lies in its simplicity and the richness of the discussion it sparks. We're not just finding a single numerical answer; we're exploring how the solution behaves as we change the parameter a. This is a core concept in many areas of mathematics, including calculus, differential equations, and linear algebra. So, stick with me as we untangle this equation and reveal its secrets!

Step-by-Step Solution

Okay, let's get our hands dirty and solve this equation step-by-step. Remember, we want to isolate x. Here's how we'll do it:

Rearrange the terms: Our first move is to gather all the terms containing x on one side of the equation. Let's subtract $(a+1)x$ from both sides: $a^2 x + (a-1) - (a+1)x = (a+1)x - (a+1)x$ This simplifies to: $a^2 x - (a+1)x + (a-1) = 0$
Factor out x: Now, we can factor out x from the terms on the left-hand side: $x(a^2 - (a+1)) + (a-1) = 0$
Simplify the coefficient of x: Let's simplify the expression inside the parentheses: $x(a^2 - a - 1) + (a-1) = 0$
Isolate the x term: Next, we'll subtract $(a-1)$ from both sides to isolate the term containing x: $x(a^2 - a - 1) = -(a-1)$ Which can also be written as: $x(a^2 - a - 1) = 1 - a$
Solve for x: Finally, to solve for x, we'll divide both sides by $(a^2 - a - 1)$ . But hold on! We need to be cautious here. Dividing by zero is a big no-no in mathematics. So, we need to consider the case where $(a^2 - a - 1) = 0$ separately. Assuming for now that $(a^2 - a - 1) eq 0$ , we can divide: $x = rac{1-a}{a^2 - a - 1}$

Now, we have a general solution for x, but it comes with a caveat. We need to investigate what happens when the denominator, $(a^2 - a - 1)$ , is equal to zero. This will lead us to some interesting special cases.

Analyzing the Special Cases

As we discovered in the previous section, the solution for x is given by $x = rac{1-a}{a^2 - a - 1}$ , provided that the denominator, $a^2 - a - 1$ , is not zero. But what happens when $a^2 - a - 1 = 0$ ? This is where things get interesting, and we need to put on our detective hats to analyze these special cases.

Case 1: $a^2 - a - 1 = 0$

First, let's find the values of a that make the denominator zero. We can do this by solving the quadratic equation $a^2 - a - 1 = 0$ . We can use the quadratic formula, which states that for an equation of the form $ax^2 + bx + c = 0$ , the solutions are given by:

$x = rac{-b ext{ ± } ext{√}(b^2 - 4ac)}{2a}$

In our case, the variable is a, and the coefficients are: a = 1, b = -1, and c = -1. Plugging these values into the quadratic formula, we get:

$a = rac{-(-1) ext{ ± } ext{√}((-1)^2 - 4(1)(-1))}{2(1)}$

$a = rac{1 ext{ \pm } ext{\sqrt}(1 + 4)}{2}$

$a = rac{1 ext{ \pm } ext{\sqrt}5}{2}$

So, we have two values of a that make the denominator zero: $a_1 = rac{1 + ext{√}5}{2}$ and $a_2 = rac{1 - ext{√}5}{2}$ . These are the famous golden ratio (φ) and its conjugate! Now, we need to investigate what happens to the original equation when a takes on these values.

Sub-cases when $a^2 - a - 1 = 0$

When $a^2 - a - 1 = 0$ , our equation $x(a^2 - a - 1) = 1 - a$ becomes $x(0) = 1 - a$ , which simplifies to $0 = 1 - a$ . This is a crucial point! It means that if $1 - a$ is also zero, then the equation becomes $0 = 0$ , which is always true. If $1 - a$ is not zero, then we have a contradiction ( $0 =$ something non-zero), meaning there's no solution.

Sub-case 1.1: If $1 - a = 0$ , then $a = 1$ . However, we already know that a must be either $rac{1 + ext{\sqrt}5}{2}$ or $rac{1 - ext{\sqrt}5}{2}$ for the denominator to be zero. So, $a = 1$ doesn't fit our condition for this case. Therefore, when $a = rac{1 + ext{\sqrt}5}{2}$ or $a = rac{1 - ext{\sqrt}5}{2}$ , and $1-a eq 0$ there are no solutions.

Case 2: $a^2 - a - 1

eq 0$

This is the case we initially considered, where we could safely divide by $a^2 - a - 1$ . In this case, the solution is simply:

$x = rac{1-a}{a^2 - a - 1}$

This gives us a unique solution for x as long as a is not equal to $rac{1 + ext{\sqrt}5}{2}$ or $rac{1 - ext{\sqrt}5}{2}$ .

Summarizing the Solutions

Alright, guys, we've journeyed through the equation $a^2 x + (a-1) = (a+1)x$ and uncovered its secrets. Let's recap what we've found:

General Solution: If $a^2 - a - 1 eq 0$ , then the equation has a unique solution given by: $x = rac{1-a}{a^2 - a - 1}$
Special Cases:
- If $a = rac{1 + ext{\sqrt}5}{2}$ or $a = rac{1 - ext{\sqrt}5}{2}$ , then $a^2 - a - 1 = 0$ . In these cases, the equation has no solution.

This comprehensive solution set paints a complete picture of how the solutions behave based on the value of the parameter a. We've seen how a seemingly simple linear equation can lead to a rich mathematical discussion involving quadratic equations, the golden ratio, and careful consideration of special cases. This is what makes math so fascinating!

Graphical Interpretation (Optional)

For those of you who are visually inclined, let's briefly touch upon how we can interpret these solutions graphically. The equation $a^2 x + (a-1) = (a+1)x$ can be rearranged to represent two lines:

Line 1: $y = (a+1)x$
Line 2: $y = a^2 x + (a-1)$

The solution to our equation corresponds to the x-coordinate of the point where these two lines intersect. The slope of Line 1 is $(a+1)$ , and the slope of Line 2 is $a^2$ . The y-intercept of Line 2 is $(a-1)$ .

When $a^2 - a - 1 eq 0$ , the lines have different slopes and will intersect at a single point, giving us a unique solution for x. This aligns with our general solution.
When $a = rac{1 + ext{\sqrt}5}{2}$ or $a = rac{1 - ext{\sqrt}5}{2}$ , the slopes might be equal, and the lines may be parallel. If the y-intercepts are different, the lines never intersect, leading to no solution. If the lines coincide (same slope and y-intercept), there would be infinitely many solutions, but in our case, we found no solutions because $1-a$ was not zero when $a^2 - a - 1 = 0$ .

This graphical perspective provides another layer of understanding to the solutions we derived algebraically.

Conclusion

So, there you have it! We've thoroughly dissected the equation $a^2 x + (a-1) = (a+1)x$ , exploring its solution set and the special cases that arise. We've seen how the value of the parameter a dictates the nature of the solutions, leading to either a unique solution or no solution at all. By carefully analyzing the equation and considering potential pitfalls like division by zero, we've gained a deeper appreciation for the nuances of solving algebraic equations.

Remember, guys, mathematics is not just about finding answers; it's about understanding the why behind the answers. It's about exploring the relationships between numbers and variables and uncovering the hidden patterns that govern the mathematical world. Keep practicing, keep questioning, and keep exploring! And who knows, maybe you'll be the one to unravel the next great mathematical mystery!