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Magazine Chapter 1: Problem 50
1鈥�54 ? Find all real solutions of the equation. $$ \sqrt[3]{4 x^{2}-4 x}=x $$
Short Answer
Expert verified
The solutions are \( x = 0 \) and \( x = 2 \).
Step by step solution
01
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.
02
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 \( 4x^2 - 4x = x^3 \).
03
Rearrange the Equation
Rearrange the equation to bring all terms to one side: \( x^3 - 4x^2 + 4x = 0 \).
04
Factor the Equation
Factor out the greatest common factor, \( x \), from the equation: \( x(x^2 - 4x + 4) = 0 \).
05
Solve Simple Factor
The equation \( x = 0 \) is one solution.
06
Factor Further
The quadratic part \( x^2 - 4x + 4 \) is a perfect square. Factor it as \( (x - 2)^2 \). The equation becomes \( x(x - 2)^2 = 0 \).
07
Solve Factored Equation
Set each factor to zero: \( x = 0 \) and \( (x - 2)^2 = 0 \). The second equation \( (x - 2)^2 = 0 \) gives \( x = 2 \).
08
List All Solutions
The complete set of solutions is \( x = 0 \) and \( x = 2 \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
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.
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.
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.
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.
