Problem 22 - Algebra and Trigonometry

Question

Show that \(100 n \leq n^{2}\) for all \(n \geq 100.\)

Step-by-Step Solution

Answer

The inequality holds for all \(n \geq 100\).

Step 1: Understand the Inequality

We need to show that the inequality \(100n \leq n^2\) holds true for all integers \(n\) that are equal to or greater than 100. This means we want to confirm that the left side of the equation is less than or equal to the right side.

Step 2: Rearrange the Inequality

Rearrange the inequality to make it easier to analyze by subtracting \(100n\) from both sides:\[ n^2 - 100n \geq 0 \] This will help us factor and solve the problem.

Step 3: Factor the Quadratic Expression

Factor the quadratic expression \(n^2 - 100n\):\[ n(n - 100) \geq 0 \] This suggests the product of \(n\) and \(n - 100\) must be greater than or equal to zero.

Step 4: Analyze the Critical Points

Identify the critical points, which are the values of \(n\) that make each factor zero: \(n = 0\) and \(n = 100\). The inequality will change sign at these critical points.

Step 5: Test the Intervals

Test the intervals created by the critical points. For \(n \geq 100\), we analyze the interval starting at \(n = 100\). Since both factors \(n\) and \(n-100\) are positive for \(n > 100\), the product \(n(n-100)\) is positive, satisfying \(n(n-100) \geq 0\).

Step 6: Confirm Satisfaction of Inequality

For \(n = 100\), plug back into the original inequality to confirm: \(100^2 \geq 100 \times 100\) which holds true as both are equal to 10000. This shows the inequality holds at \(n = 100\) and for all \(n > 100\).

Key Concepts

Factoring Quadratics

When dealing with quadratic inequalities, one fundamental approach is factoring the involved quadratic expression. In this exercise, our quadratic expression is rearranged to: \[ n^2 - 100n \].
A quadratic can often be expressed as the product of two binomials. This is what happened when we factored the expression:

• We identified the common term, \(n\), in both parts of \(n(n - 100)\).
• This results in the equation: \[ n(n - 100) \geq 0 \].

The key here is that factoring simplifies the inequality, making it easier to identify where the inequality changes sign.

Critical Points

Critical points are the values of the variable where each factor of the expression becomes zero. These help identify where the inequality may change from true to false or vice versa. For our factored inequality \[ n(n - 100) \geq 0 \], the critical points are derived from setting each factor equal to zero.

• If \(n = 0\), the factor \(n\) is zero.
• If \(n = 100\), the factor \(n - 100\) is zero.

These critical points divide the number line into intervals, which are crucial for further analysis.

Interval Testing

Once we have found our critical points, the next step is to test intervals defined by these points. The critical points, \(n = 0\) and \(n = 100\), divide the number line into intervals \((-\infty, 0)\), \((0, 100)\), and \((100, \infty)\).
Interval testing involves checking a single point from each interval into the inequality to see where it holds true.

• For \(n > 100\), choose a point like \(n = 101\). Here, both \(n\) and \(n - 100\) are positive, ensuring \(n(n-100) \geq 0\).
• For \(n = 100\), check directly: \[ 100(100 - 100) = 0 \].
• The interval \((0, 100)\) does not need testing for \(n \geq 100\). Testing confirms where the inequality is satisfied.

Analyzing Inequalities

After performing interval testing, we are ready to analyze our results to confirm the validity of the inequality across the considered range.
The inequality \[ n^2 - 100n \geq 0 \] is satisfied when both factors are either both positive or both zero. The analysis involves:

• Verifying the product \(n(n - 100)\) is non-negative for the specific range \( n \geq 100 \).
• Ensuring the result holds true at key points like \( n = 100 \), where originally posed inequality \( 100^2 \geq 100 \times 100 \) equals \(10000\).

Through this comprehensive analysis, the inequality is confirmed valid. Understanding the behavior of quadratic expressions within defined intervals allows us to solve a myriad of real-world and mathematical problems effectively.