Problem 1
Question
1–54 ? Find all real solutions of the equation. $$ x^{3}=16 x $$
Step-by-Step Solution
Verified Answer
The real solutions are \(x = 0, x = 4, \) and \(x = -4\).
1Step 1: Set Equation to Zero
First, we need to rewrite the equation into the standard form by bringing all terms to one side. Start with the equation:\[ x^3 = 16x \]Subtract \(16x\) from both sides:\[ x^3 - 16x = 0 \]
2Step 2: Factor the Equation
Now, factor the common factor from the equation. Notice that both terms have an \(x\) in common:\[ x(x^2 - 16) = 0 \]
3Step 3: Solve for Zero Factor Property
Apply the Zero Factor Property, which states that if the product of factors is zero, at least one of the factors must be zero. Thus, set each factor equal to zero:1. \( x = 0 \)2. \( x^2 - 16 = 0 \)
4Step 4: Solve the Quadratic Equation
Solve the quadratic equation \(x^2 - 16 = 0\) by adding 16 to both sides:\[ x^2 = 16 \]Now, take the square root of both sides:\[ x = \pm \sqrt{16} \]This simplifies to:\[ x = \pm 4 \]
5Step 5: Write Down All Solutions
Combine all solutions from Step 3 and Step 4. The solutions to the equation are:\[ x = 0, \, x = 4, \, x = -4 \]
Key Concepts
Factoring EquationsZero Factor PropertyQuadratic Equations
Factoring Equations
Factoring equations is a method used to simplify and solve equations. In simple terms, factoring is the process of breaking down an equation into simpler components called factors. This can make solving the original equation much easier. It's similar to finding the prime factors of a number – instead of dealing with the complex number directly, you work with its more manageable parts.
Consider the equation given in the exercise: \[ x^3 - 16x = 0 \]This equation can be factored by identifying a common term in each of the equation's parts. Here, you can see that both terms, \(x^3\) and \(-16x\), include an \(x\), so you can factor out \(x\) as follows:\[ x(x^2 - 16) = 0 \] By breaking it into these factors, it becomes simpler to solve the problem. Factoring thus turns a complicated equation into a product of simpler factors, each of which can potentially be solved individually.
Consider the equation given in the exercise: \[ x^3 - 16x = 0 \]This equation can be factored by identifying a common term in each of the equation's parts. Here, you can see that both terms, \(x^3\) and \(-16x\), include an \(x\), so you can factor out \(x\) as follows:\[ x(x^2 - 16) = 0 \] By breaking it into these factors, it becomes simpler to solve the problem. Factoring thus turns a complicated equation into a product of simpler factors, each of which can potentially be solved individually.
Zero Factor Property
The Zero Factor Property (or Zero Product Property) is an essential concept in algebra, which makes it possible to find solutions to equations once they have been factored. This property states that if the product of multiple factors equals zero, at least one of the factors must be zero. In mathematical terms, if \( a \times b = 0 \), then either \( a = 0 \), \( b = 0 \), or both could be zero.
In our exercise:\[ x(x^2 - 16) = 0 \]The Zero Factor Property allows us to set each factor equal to zero:
In our exercise:\[ x(x^2 - 16) = 0 \]The Zero Factor Property allows us to set each factor equal to zero:
- \( x = 0 \)
- \( x^2 - 16 = 0 \)
Quadratic Equations
Quadratic equations are polynomial equations of the second degree, generally written in the form \( ax^2 + bx + c = 0 \). Solving these equations involves finding the values of \( x \) that make the equation true. These solutions can be real or complex numbers.
In the exercise, after applying the Zero Factor Property, we are left with:\[ x^2 - 16 = 0 \]This is an example of a quadratic equation that can be solved using basic algebraic manipulation. First, add 16 to both sides:\[ x^2 = 16 \]Then, by taking the square root of both sides, we find that:\[ x = \pm 4 \]Thus, the quadratic yields two solutions: \( x = 4 \) and \( x = -4 \).
Quadratic equations often have two solutions, which could be the same or different, and may be real or complex. In this exercise, you found all possible real solutions. Understanding how to manipulate and solve quadratic equations is crucial for dealing with a wide range of mathematical problems.
In the exercise, after applying the Zero Factor Property, we are left with:\[ x^2 - 16 = 0 \]This is an example of a quadratic equation that can be solved using basic algebraic manipulation. First, add 16 to both sides:\[ x^2 = 16 \]Then, by taking the square root of both sides, we find that:\[ x = \pm 4 \]Thus, the quadratic yields two solutions: \( x = 4 \) and \( x = -4 \).
Quadratic equations often have two solutions, which could be the same or different, and may be real or complex. In this exercise, you found all possible real solutions. Understanding how to manipulate and solve quadratic equations is crucial for dealing with a wide range of mathematical problems.
Other exercises in this chapter
Problem 1
\(1-8=\) Let \(S=\left\\{-2,-1,0, \frac{1}{2}, 1, \sqrt{2}, 2,4\right\\} .\) Determine which elements of \(S\) satisfy the inequality. $$ x-3>0 $$
View solution Problem 1
Find the real and imaginary parts of the complex number. $$ 5-7 i $$
View solution Problem 1
Solve the equation by factoring. \(x^{2}+x-12=0\)
View solution Problem 1
\(1-8\) Determine whether the given value is a solution of the equation. $$ \begin{array}{l}{4 x+7=9 x-3} \\ {\text { (a) } x=-2} & {\text { (b) } x=2}\end{arra
View solution