Problem 24
Question
Show that \(100 n \leq n^{2}\) for all \(n \geq 100\)
Step-by-Step Solution
Verified Answer
The inequality \( 100n \leq n^2 \) holds true for all \( n \geq 100 \).
1Step 1: Understand the Inequality
We need to show that for all values of \( n \) greater than or equal to 100, the inequality \( 100n \leq n^2 \) holds. This means that for these values of \( n \), multiplying \( n \) by 100 will always be less than or equal to \( n \) squared.
2Step 2: Rearrange the Inequality
Rearrange the given inequality \( 100n \leq n^2 \) into a more workable form by subtracting \( 100n \) from both sides. This results in the inequality: \( n^2 - 100n \geq 0 \).
3Step 3: Factorize the Expression
Factor the quadratic expression \( n^2 - 100n \) to simplify further analysis. This can be factored as: \( n(n - 100) \geq 0 \).
4Step 4: Analyze the Factored Inequality
Analyze the inequality \( n(n - 100) \geq 0 \). The product of two numbers is non-negative when both are non-negative or both are non-positive. For \( n \geq 100 \), both \( n \) and \( (n - 100) \) are non-negative.
5Step 5: Conclusion
Since both factors are non-negative for \( n \geq 100 \), \( n(n - 100) \geq 0 \) is true. Thus, the inequality \( 100n \leq n^2 \) holds for all \( n \geq 100 \).
Key Concepts
Quadratic InequalityFactoring ExpressionsAlgebraic Analysis
Quadratic Inequality
A quadratic inequality involves a quadratic expression and an inequality symbol, such as \( \leq \), \( \geq \), \( < \), or \( > \). In our case, we want to demonstrate that the inequality \( 100n \leq n^2 \) holds for all \( n \geq 100 \). To tackle such inequalities, we first rearrange terms in a manner that resembles a standard quadratic form (\( ax^2 + bx + c \geq 0 \)).
\( n^2 - 100n \geq 0 \) is the rearranged inequality and a common type seen in algebra, where you later determine the intervals in which it holds true.
Quadratic inequalities can often be solved by determining when the quadratic product changes signs. **The critical part is understanding when the product is positive or zero.**
This is usually established through factoring and testing sign changes across critical points derived from the equation's roots.
\( n^2 - 100n \geq 0 \) is the rearranged inequality and a common type seen in algebra, where you later determine the intervals in which it holds true.
Quadratic inequalities can often be solved by determining when the quadratic product changes signs. **The critical part is understanding when the product is positive or zero.**
This is usually established through factoring and testing sign changes across critical points derived from the equation's roots.
Factoring Expressions
Factoring is a crucial algebraic method used to simplify quadratic expressions which transforms them into a product of simpler expressions. In the case of \( n^2 - 100n \), factoring gives us \( n(n - 100) \).
This reveals the roots of the equation: 0 and 100, since one or both products become zero at these points.
This reveals the roots of the equation: 0 and 100, since one or both products become zero at these points.
- Factoring helps us establish potential intervals of positivity or negativity in a quadratic inequality.
- It allows a straightforward application of the zero-product property: a product is zero only if one or more of its factors are zero.
Algebraic Analysis
Algebraic analysis involves examining expressions or equations to derive meaningful conclusions about their behaviour across different values. When dealing with inequalities like \( n(n - 100) \geq 0 \), this analysis determines where the expression is positive.
Nearly, we check if both factors of the expression are positive or negative. With \( n(n - 100) \):
**This concept extends to other algebraic issues, predicting whether solutions will work universally or only in certain conditions.**
Nearly, we check if both factors of the expression are positive or negative. With \( n(n - 100) \):
- For \( n \geq 100 \), **both factors are non-negative**, ensuring the product \( \geq 0 \).
- Strategic selection of test points, usually around the roots, quickly confirms sign consistency within intervals.
**This concept extends to other algebraic issues, predicting whether solutions will work universally or only in certain conditions.**
Other exercises in this chapter
Problem 24
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