Problem 50 - Algebra and Trigonometry

Question

1–54 ? Find all real solutions of the equation. $$ \sqrt[3]{4 x^{2}-4 x}=x $$

Step-by-Step Solution

Answer

The solutions are \( x = 0 \) and \( x = 2 \).

Step 1: Understand the Equation

The given equation is \( \sqrt[3]{4x^2 - 4x} = x \). This means that we are looking for the values of \( x \) that make both sides equal.

Step 2: Cube Both Sides

To remove the cube root, cube both sides of the equation: \( (\sqrt[3]{4x^2 - 4x})^3 = x^3 \). This simplifies to \( 4x^2 - 4x = x^3 \).

Step 3: Rearrange the Equation

Rearrange the equation to bring all terms to one side: \( x^3 - 4x^2 + 4x = 0 \).

Step 4: Factor the Equation

Factor out the greatest common factor, \( x \), from the equation: \( x(x^2 - 4x + 4) = 0 \).

Step 5: Solve Simple Factor

The equation \( x = 0 \) is one solution.

Step 6: Factor Further

The quadratic part \( x^2 - 4x + 4 \) is a perfect square. Factor it as \( (x - 2)^2 \). The equation becomes \( x(x - 2)^2 = 0 \).

Step 7: Solve Factored Equation

Set each factor to zero: \( x = 0 \) and \( (x - 2)^2 = 0 \). The second equation \( (x - 2)^2 = 0 \) gives \( x = 2 \).

Step 8: List All Solutions

The complete set of solutions is \( x = 0 \) and \( x = 2 \).

Key Concepts

Real Solutions

In algebra, a real solution is a value that satisfies an equation. It means the solution is a real number rather than an imaginary one. Real numbers include integers, fractions, and decimals, and they can be positive or negative.
When solving the equation \( \sqrt[3]{4x^2 - 4x} = x \), the aim is to find values of \( x \) that make this equation true by balancing both sides. These values are referred to as real solutions.

• Identify the expressions involved, here being a cube root expression and a simple \( x \).
• Manipulate the equation in such a way that arithmetic properties can simplify it.
• Ensure that the solutions are within the set of real numbers.

Real solutions are important in mathematics because they represent tangible, countable answers that we can verify through substitution.

Factoring Polynomials

Factoring polynomials is a critical method in algebra that involves expressing a polynomial as a product of its factors. Factors are simpler expressions that multiply together to give the original polynomial.
In the equation \( x^3 - 4x^2 + 4x = 0 \), one can see that the expression can be simplified by factoring.

• Identify a common factor across all terms; in this case, \( x \).
• Factor this common factor out: \( x(x^2 - 4x + 4) = 0 \).
• Look for easy-to-spot special factorizations, such as squares.

After factoring, it’s easier to find the values of \( x \) that satisfy the equation. Factoring is an essential skill as it simplifies complex equations, allowing for straightforward solving.

Cube Roots

The cube root is the inverse operation of raising a number to the power of three. In the equation \( \sqrt[3]{4x^2 - 4x} = x \), the cube root symbol \( \sqrt[3]{} \) implies that we need to find a number, that when cubed, equals to the expression inside the root.

• To deal with the cube root, we cube both sides to simplify the equation: \( (\sqrt[3]{4x^2 - 4x})^3 = x^3 \).
• This operation cancels the cube root, resulting in \( 4x^2 - 4x = x^3 \).
• Cubing eliminates the complexity introduced by the radical.

Understanding cube roots and their properties is vital, especially when dealing with polynomial equations. It provides a direct way to tackle equations that involve cubic terms.

Quadratic Equations

A quadratic equation is any equation that can be rearranged into the form \( ax^2 + bx + c = 0 \). This forms part of our problem when we get the expression \( x^2 - 4x + 4 \) after factoring the polynomial.
Quadratic equations can often be solved by recognizing it as a perfect square, using the quadratic formula, or factoring.

• In our exercise, \( x^2 - 4x + 4 \) can be observed as a perfect square: \( (x - 2)^2 \).
• By setting this factor to zero, \( (x - 2)^2 = 0 \), we derive \( x = 2 \).
• Quadratic equations are versatile and appear frequently in many mathematical contexts.

Understanding how to handle quadratics opens a wide range of problem-solving techniques, essential in both learning and practical applications of algebra.