Problem 73 - Algebra for College Students

Question

Solve each inequality. State the solution set using interval notation when possible. \(\frac{1}{2} x^{2} \geq 4-x\)

Step-by-Step Solution

Answer

The solution set is \(x \leq -4 \cup x \geq 2\) in interval notation.

Step 1 - Move all terms to one side

Rewrite the inequality \(\frac{1}{2} x^{2} \geq 4-x\) by moving all terms to one side: \(\frac{1}{2} x^2 + x - 4 \geq 0\)

Step 2 - Multiply by 2

To clear the fraction, multiply each term by 2: \(x^2 + 2x - 8 \geq 0\)

Step 3 - Factor the quadratic

Factor the quadratic expression \(x^2 + 2x - 8\) to simplify it. It factors to \((x + 4)(x - 2)\). Therefore, the inequality becomes \((x + 4)(x - 2) \geq 0\).

Step 4 - Find critical points

Set each factor equal to zero to determine the critical points: \(x + 4 = 0\) gives \(x = -4\) and \(x - 2 = 0\) gives \(x = 2\). These points divide the number line into intervals.

Step 5 - Test intervals

Test each interval determined by the critical points in the inequality \((x + 4)(x - 2) \geq 0\)\.- For \x < -4\ (e.g., \x = -5): \((-5 + 4)(-5 - 2) = (-1)(-7) = 7 \geq 0\)- For \-4 < x < 2\ (e.g., \x = 0): \((0 + 4)(0 - 2) = 4(-2) = -8 \lt 0\)- For \x > 2\ (e.g., \x = 3): \((3 + 4)(3 - 2) = 7(1) = 7 \geq 0\)

Step 6 - Solution set

Summarize the results from the interval testing. The solution set where the inequality holds is \(x \leq -4 \cup x \geq 2\).

Key Concepts

quadratic inequalities

When we talk about quadratic inequalities, we deal with expressions that include a quadratic term (something with an x^2). Unlike basic linear inequalities, quadratic inequalities can result in solution sets that are ranges (intervals) rather than just single points. Here, we start with an inequality: \(\frac{1}{2} x^{2} \geq 4-x\). 
First, we rearrange everything to one side to get it into the standard quadratic form. This helps us clearly see the quadratic expression we will work with. The standard approach is to get: \( \frac{1}{2} x^2 + x - 4 \geq 0 \).
To simplify, we often clear any fractions by multiplying through. This leads to: \( x^2 + 2x - 8 \geq 0 \). This form can now be easily analyzed. The key here is to transform the inequality into a form we can factor.

factoring quadratics

Factoring quadratics involves breaking down a quadratic equation into simpler components. Specifically, we look for two binomials whose product gives us the original quadratic equation.
For the expression \( x^2 + 2x - 8 \), we need to find two numbers that multiply to \( -8 \) (the constant term) and add up to \( 2 \) (the coefficient of the x term).
Upon inspection, these numbers are \( 4 \) and \( -2 \), because \( 4 \times -2 = -8 \) and \( 4 + (-2) = 2 \).
Hence, the factored form of \( x^2 + 2x - 8 \) is: \( (x + 4)(x - 2) \).
Now our inequality looks like \( (x + 4)(x - 2) \geq 0 \). Factoring makes it much easier to analyze where the inequality holds true by finding the critical points.

interval notation

Interval notation is a method of writing down sets of numbers that are solutions to inequalities. Instead of listing every solution, interval notation allows us to describe ranges compactly.
Based on our inequality, we identify the critical points by setting each factor equal to zero:
\( x + 4 = 0 \) leads to \( x = -4 \) and \( x - 2 = 0 \), which results in \( x = 2 \).
These points divide the number line into distinct intervals.

• For \( x < -4 \) or to the left of \( -4 \)
• For \( -4 \leq x \leq 2 \)
• For \( x > 2 \) or to the right of \( 2 \)

Interval notation then helps us write the solution set for the inequality. From our testing, we determine that the solution set is: \( x \leq -4 \cup x \geq 2 \), which means that \( x \) is either less than or equal to \( -4 \) or greater than or equal to \( 2 \).

testing intervals

Testing intervals involves checking whether the inequality holds true within specific ranges on the number line. Using the critical points, we split the number line into intervals, then select a test point from each interval.

• Test \( x < -4 \): Choose \( x = -5 \). \( ( -5 + 4)( -5 - 2) = (-1)(-7) = 7 \geq 0 \)
• Test \( -4 < x < 2 \): Choose \( x = 0 \). \( (0 + 4)( 0 - 2) = 4(-2) = -8 \lt 0 \)
• Test \( x > 2 \): Choose \( x = 3 \). \( (3 + 4)( 3 - 2) = 7(1) = 7 \geq 0 \)

By evaluating the signs of these products, we can determine where the original inequality holds true.
For our example, the inequality holds for \( x \leq -4 \) and \( x \geq 2 \).