Problem 14 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ (x-2)^{5}-9(x-2)^{3}=0 $$

Step-by-Step Solution

Answer

The real solutions are \(x = 2\), \(x = 5\), and \(x = -1\).

Step 1: Identify a substitution

Notice that the equation is in terms of \((x-2)\). Let's make a substitution: let \(y = (x-2)\). Thus, the equation becomes \(y^5 - 9y^3 = 0\).

Step 2: Factor the equation

The equation \(y^5 - 9y^3 = 0\) can be factored by taking \(y^3\) common: \(y^3(y^2 - 9) = 0\).

Step 3: Solve the factored equation

Set each factor to zero: \(y^3 = 0\) and \(y^2 - 9 = 0\).

Step 4: Solve \(y^3 = 0\)

Solving \(y^3 = 0\) gives \(y = 0\).

Step 5: Solve \(y^2 - 9 = 0\)

Solving \(y^2 - 9 = 0\) gives \(y = 3\) and \(y = -3\) by taking the square root of both sides.

Step 6: Resubstitute \(y\) back to \(x\)

Recall \(y = x - 2\). Solve for \(x\) in each case: - If \(y = 0\), then \(x - 2 = 0\), so \(x = 2\).- If \(y = 3\), then \(x - 2 = 3\), so \(x = 5\).- If \(y = -3\), then \(x - 2 = -3\), so \(x = -1\).

Step 7: Write the solutions

The solutions for the equation are \(x = 2\), \(x = 5\), and \(x = -1\).

Key Concepts

Algebraic substitution

In algebra, substitution is a technique used to simplify equations, making them easier to solve. When we deal with equations involving complex expressions, substitution can help to transform the problem into a more manageable form. For example, in the exercise given, the equation
\[(x-2)^5 - 9(x-2)^3 = 0\]is primarily expressed in terms of \(x-2\). To simplify the process, we chose to substitute \(y = (x-2)\), thereby reducing the equation to \[ y^5 - 9y^3 = 0 \].
This substitution refocuses the equation in terms of a single variable \(y\), making subsequent operations like factoring more straightforward.

• Transforms the complex expressions into simpler forms.
• Makes equations more manageable and solvable.
• Facilitates easier application of other algebraic methods.

Once solved, we can reverse the substitution to find the solutions of the original equation, providing the real roots based on this transformation. This is particularly useful when handling higher-degree polynomials, as it allows us to break them into more understandable parts.

Factoring polynomials

Factoring is a pivotal concept in solving polynomial equations. When faced with an equation like \[ y^5 - 9y^3 = 0 \], factoring involves breaking the expression into a product of simpler polynomials that can be individually solved.In the example above, we notice each term shares a common factor of \(y^3\). By factoring, the equation becomes:\[ y^3(y^2 - 9) = 0 \].
This reveals two distinct components that can each be zero, allowing us to set up separate equations: \(y^3 = 0\) and \(y^2 - 9 = 0\).

• Simplifies polynomial equations into manageable components.
• Uncovers the roots by breaking down complex equations.
• Utilizes common factors to create products of simpler polynomials.

Understanding factoring is crucial as it highlights methods to find real solutions through simplification. Moreover, once the distinct linear (or quadratic) factors are identified, they represent paths to finding the specific values of a variable that make the equation true.

Real solutions

In algebra, finding real solutions pertains to identifying actual values of variables that satisfy an equation in real numbers.
When the polynomial equation is set up, as in \[ y^3(y^2 - 9) = 0 \],we seek real values for \(y\) as roots of the equation. Solving \(y^3 = 0\) provides \(y = 0\), while \(y^2 - 9 = 0\) leads to solutions \(y = 3\) and \(y = -3\) by utilizing square roots.

• Solutions belong to the set of real numbers, as opposed to imaginary.
• Signify values that make the equation true upon substitution.
• May include multiple roots, including negative, zero, or positive values.

Real solutions for the values \(y\), such as 0, 3, and -3, need re-substitution back to find the corresponding \(x\) values like in the original problem, providing a clear understanding of where each solution fits within the context of the original equation.

Algebra

Algebra serves as the foundation of solving equations, involving variables, operations, and understanding relationships between these mathematical expressions.
In our exercise, algebra guides us through steps from substitution and factoring to identifying solutions. Working through \[(x-2)^5 - 9(x-2)^3 = 0\],and breaking it down through algebraic methods to solve for \(x\), illustrates the logical structure and techniques we use in algebra.

• Substitution and transformation of equations into simpler forms.
• Utilizing operations like factoring to identify potential solutions.
• Applying problem-solving skills to transition between variables and uncover true values.

Algebra not only simplifies complex equations to a feasible level but also facilitates understanding through logical steps and operations that progressively bring out solutions in a structured manner. It aids in solving multi-step problems involving different algebraic operations and transformations, leading us to accurate real-world solutions.