Problem 54 - Precalculus: Functions and Graphs

Question

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$x(x-3) \leq 10$$

Step-by-Step Solution

Answer

The solution to the inequality is \([-2, 5]\).

Step 1: Expand the Equation

First, expand the expression on the left side. Starting with the equation \( x(x-3) = x^2 - 3x \), rewrite the inequality as \( x^2 - 3x \leq 10 \).

Step 2: Rearrange the Inequality

Bring all terms to one side to set the inequality to zero: \( x^2 - 3x - 10 \leq 0 \). Now we have a quadratic inequality.

Step 3: Factor the Quadratic Expression

Next, we need to factor the quadratic: \( x^2 - 3x - 10 \). Find two numbers that multiply to \(-10\) and add to \(-3\). These numbers are \(-5\) and \(2\). Thus, the factors are \((x-5)(x+2)\).

Step 4: Identify Critical Points

Set each factor equal to zero to find the critical points, solving \(x-5 = 0\) and \(x+2 = 0\), which yields \(x = 5\) and \(x = -2\). These critical points divide the number line into intervals.

Step 5: Test Intervals

Test the intervals \((-\infty, -2)\), \((-2, 5)\), and \((5, \infty)\) to see where the inequality \((x-5)(x+2) \leq 0\) holds true. Pick test points: \(-3\), \(0\), and \(6\). Plug them into the factored inequality: - For \(x = -3\): \((-3 - 5)(-3 + 2) = 8(1) = 8 > 0\). - For \(x = 0\): \((0 - 5)(0 + 2) = (-5)(2) = -10 < 0\).- For \(x = 6\): \((6 - 5)(6 + 2) = 1(8) = 8 > 0\). The inequality holds in the interval \((-2, 5)\).

Step 6: Include Critical Points

Since the inequality is \(\leq 0\), include the critical points \(x = -2\) and \(x = 5\). Therefore, the solution is \([-2, 5]\).

Key Concepts

Factoring Quadratics

In solving quadratic inequalities, one of the key steps is to factor the quadratic expression. This transforms the inequality into a product of two binomials set to zero or less. In our example, we started with the inequality \(x^2 - 3x - 10 \leq 0\).
The goal is to express this quadratic in its factored form.To do this, we identify two numbers that both add up to the coefficient of \(x\) and multiply to the constant term. Here, we needed numbers that multiply to \(-10\) and add to \(-3\).
After a bit of searching, we find these numbers are \(-5\) and \(2\). These allow us to rewrite the quadratic as \((x-5)(x+2)\). This makes the inequality more manageable and sets us up for the next steps.

Critical Points

Once the quadratic expression is factored, the next crucial part of solving the inequality is to find the critical points. These are the values of \(x\) that make each factor equal to zero.
They essentially divide the number line into intervals where the inequality may hold true or false.In our example, the factors are \((x-5)\) and \((x+2)\). Set each of these to zero:

• \(x-5=0\) gives \(x=5\)
• \(x+2=0\) gives \(x=-2\)

These critical points are pivotal in interval testing, as they mark changes in the sign of the expression. Remember, inequalities can change from true to false as \(x\) crosses these points.

Interval Testing

Interval testing involves selecting test points from the intervals determined by the critical points. These intervals help determine where the inequality holds true.
The critical points \(-2\) and \(5\) split the number line into three intervals: \(( -\infty, -2)\), \((-2, 5)\), and \((5, \infty)\).From each interval, choose a test point:

• In \(( -\infty, -2)\), try \(x = -3\)
• In \((-2, 5)\), try \(x = 0\)
• In \((5, \infty)\), try \(x = 6\)

Plug these points into the factored inequality \((x-5)(x+2)\) and evaluate.This process reveals where the inequality \((x-5)(x+2) \leq 0\) is true, essential for finding the solution interval.

Solution Intervals

After testing intervals, the solution to the inequality emerges. Based on the test results, evaluate where the inequality expression is non-positive, which means \(\leq 0\).
In our case, the inequality \((x-5)(x+2) \leq 0\) held true for \((-2, 5)\) but not for other intervals.Additionally, since the inequality is "less than or equal to," we include the critical points \(-2\) and \(5\) themselves.
Therefore, the solution interval is \([-2, 5]\). This interval includes both ends where the expression equals zero, as well as the region between them where it stays less than zero.
It is crucial to express solutions as intervals for clarity and correct mathematical representation.