Problem 7
Question
For Problems 1 through 7, give exact answers, not numerical approximations. Solve: \(\pi^{2} x^{3}=\pi x^{2}\)
Step-by-Step Solution
Verified Answer
The solution to the equation is \(x = \frac{1}{\pi}\)
1Step 1: Factor out the Common Elements
As the base \(\pi\) is a common factor, this can be factored out to simplify the equation. Given \(\pi^{2} x^{3} = \pi x^{2}\), we can divide both sides of the equation by \(\pi\) to get \(\pi x^{3} = x^{2}\)
2Step 2: Isolate the Unknown x
Our ultimate goal is to express x in terms of other constants. Therefore, we need to isolate x. Dividing both sides of the equation by \(x^{2}\), gives us \(\pi x = 1\)
3Step 3: Solve for x
Solving for x thus gives us \(x = \frac{1}{\pi}\)
Key Concepts
Factoring Common ElementsIsolating VariablesExact Solutions in Algebra
Factoring Common Elements
Understanding how to factor common elements in algebraic equations is crucial in simplifying the problem and moving towards a solution. Factoring involves identifying terms that share a common variable, number, or expression, and then rewriting the equation in a simpler form by removing these shared components.
In the exercise \(\pi^{2} x^{3}=\pi x^{2}\), the number \(\pi\) is a common element appearing in both terms of the equation. To simplify, we apply the factoring principle by dividing both sides by this common base \(\pi\), which reduces the equation's complexity. It's like unraveling a knot—once the common thread is pulled out, what remains is easier to work with.
Factoring is not limited to constants like \(\pi\); it can also be applied to variables. If you see the same variable raised to a power on both sides of an equation, that can often be factored out to simplify the process of finding a solution.
In the exercise \(\pi^{2} x^{3}=\pi x^{2}\), the number \(\pi\) is a common element appearing in both terms of the equation. To simplify, we apply the factoring principle by dividing both sides by this common base \(\pi\), which reduces the equation's complexity. It's like unraveling a knot—once the common thread is pulled out, what remains is easier to work with.
Factoring is not limited to constants like \(\pi\); it can also be applied to variables. If you see the same variable raised to a power on both sides of an equation, that can often be factored out to simplify the process of finding a solution.
Isolating Variables
Isolating variables is a fundamental technique in algebra that involves rearranging the equation to have the unknown variable on one side and everything else on the other side. This method allows us to clearly define the value of the unknown. Once other elements are simplified or factored out, the focus shifts to obtaining the value of the variable in question.
In our example, after factoring out \(\pi\), we are left with \(\pi x^{3} = x^{2}\). To isolate \(x\), we aim to have it appear just once on one side of the equation. By dividing both terms by \(x^{2}\), \(x\) is successfully isolated, resulting in \(\pi x = 1\). This clear separation makes solving for the variable more straightforward because it is now in a format that can readily be manipulated to find its exact value.
In our example, after factoring out \(\pi\), we are left with \(\pi x^{3} = x^{2}\). To isolate \(x\), we aim to have it appear just once on one side of the equation. By dividing both terms by \(x^{2}\), \(x\) is successfully isolated, resulting in \(\pi x = 1\). This clear separation makes solving for the variable more straightforward because it is now in a format that can readily be manipulated to find its exact value.
Exact Solutions in Algebra
Finding exact solutions in algebra means determining precise and accurate values for unknown variables without resorting to estimations or numerical approximations. Algebra often deals with abstract concepts, but the solutions can be concrete and exact, which is important for accuracy in mathematical and real-world applications.
To reach an exact solution, we must perform algebraic operations precisely. In the exercise, after isolating \(x\), the equation \(\pi x = 1\) guides us toward the exact value of \(x\). By dividing both sides by \(\pi\), we find that \(x = \frac{1}{\pi}\). This value is the definitive answer and represents an exact solution.
Emphasizing the concept of exact solutions encourages students to seek precise answers in algebra, establishing a solid foundation for mathematical rigor and discipline.
To reach an exact solution, we must perform algebraic operations precisely. In the exercise, after isolating \(x\), the equation \(\pi x = 1\) guides us toward the exact value of \(x\). By dividing both sides by \(\pi\), we find that \(x = \frac{1}{\pi}\). This value is the definitive answer and represents an exact solution.
Emphasizing the concept of exact solutions encourages students to seek precise answers in algebra, establishing a solid foundation for mathematical rigor and discipline.
Other exercises in this chapter
Problem 7
The velocity of an object is given in miles per hour by \(v(t)=2 t^{5}-6 t^{3}+2 t^{2}+1\) over the time interval \(-2 \leq t \leq 2\), where \(t\) is measured
View solution Problem 7
Let \(h(t)\) denote the height of a rocketship \(t\) seconds after takeoff. (a) Express the average rate of change of height of the rocket betweeen 2 and \(2.01
View solution Problem 7
Solve these inequalities and explain your answers: Think carefully. (a) \(|3 x-4|>-4\) (b) \(|3 x-4|>0\) (c) \(|3 x-4|
View solution Problem 8
The displacement of an object is given by \(d(t)=2 t^{5}-6 t^{3}+2 t^{2}+1\) miles over the time interval \(-2 \leq t \leq 2\) where \(t\) is measured in hours.
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