Problem 66
Question
The given equation involves a power of the variable. Find all real solutions of the equation. \((x-1)^{3}+8=0\)
Step-by-Step Solution
Verified Answer
The real solution is x = -1.
1Step 1: Isolate the Cubed Term
Start by isolating the \((x-1)^3\) term in the equation. To do this, subtract 8 from both sides to get \((x-1)^3 = -8\).
2Step 2: Take the Cube Root
To solve for \(x\), take the cube root of both sides of the equation. The cube root of \((x-1)^3\) is \(x-1\), and the cube root of \(-8\) is \(-2\). Therefore, \(x-1 = -2\).
3Step 3: Solve for x
Add 1 to both sides of the equation obtained in Step 2 to solve for \(x\). This results in \(x = -2 + 1\), which simplifies to \(x = -1\).
Key Concepts
Real SolutionsCube RootEquation SolvingIsolation of Terms
Real Solutions
A real solution in equations refers to results that can be found on the real number line. These solutions do not involve imaginary or complex numbers.
In the context of the given cubic equation \((x-1)^{3}+8=0\), we have found that the real solution for \(x\) is \(-1\).
This is because our solution set doesn't require any complex number arithmetic or imaginary units, making it a real solution.
In the context of the given cubic equation \((x-1)^{3}+8=0\), we have found that the real solution for \(x\) is \(-1\).
This is because our solution set doesn't require any complex number arithmetic or imaginary units, making it a real solution.
- Real solutions are practical and applicable in most everyday scenarios.
- They are found and represented on a number line between negative infinity and positive infinity.
Cube Root
Cube roots are fundamentals in solving cubic equations. To understand the cube root, note that it reverses the cubing process.
For instance, the cube of \(x\) is \(x^3\), and the cube root \(\sqrt[3]{x^3}\) returns \(x\).
In our problem, we encountered \((x-1)^3 = -8\). By taking the cube root of both sides:
For instance, the cube of \(x\) is \(x^3\), and the cube root \(\sqrt[3]{x^3}\) returns \(x\).
In our problem, we encountered \((x-1)^3 = -8\). By taking the cube root of both sides:
- The cube root of \((x-1)^3\) is \(x-1\).
- The cube root of \(-8\) is \(-2\), since \(-2 \cdot -2 \cdot -2 = -8\).
Equation Solving
Equation solving involves finding the values of \(x\) that satisfy the given equation. Every equation has specific techniques or rules that depend on the equation form.
This understanding is key to approaching equations like the cubic one, \((x-1)^3+8=0\).
This understanding is key to approaching equations like the cubic one, \((x-1)^3+8=0\).
- The first step is identifying the operations applied to the variable; in this case, cubing and adding are key operations.
- Each step strategically aims to reverse these operations to unravel \(x\).
- Simple arithmetic, like additions and subtractions, plays a crucial role in isolating or simplifying terms.
Isolation of Terms
Isolation of terms in equations refers to the process of rearranging the equation to get the variable alone on one side. This is often the first strategic move in solving equations.
For example, in our equation:
This method is crucial in solving not just cubic equations but any equation with multiple terms and operations involved.
For example, in our equation:
- Start by isolating \((x-1)^3\) on one side, by subtracting \(8\) from both sides resulting in \((x-1)^3 = -8\).
- This move simplifies the equation and makes solving for \(x\) more straightforward.
This method is crucial in solving not just cubic equations but any equation with multiple terms and operations involved.
Other exercises in this chapter
Problem 66
Use the discriminant to determine the number of real solutions of the equation. Do not solve the equation. $$ x^{2}=6 x-9 $$
View solution Problem 66
\(61-70\) . Find all solutions, real and complex, of the equation. $$ x^{3}+3 x^{2}+9 x+27=0 $$
View solution Problem 67
Find all solutions of the equation, and express them in the form \(a+b i\) $$ t+3+\frac{3}{t}=0 $$
View solution Problem 67
Solve the nonlinear inequality. Express the solution using interval notation and graph the solution set. $$ \frac{6}{x-1}-\frac{6}{x} \geq 1 $$
View solution