Problem 53 - Precalculus: Functions and Graphs

Question

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$10 x^{2}+11 x>6$$

Step-by-Step Solution

Answer

\((-\infty, -\frac{3}{2}) \cup (\frac{2}{5}, \infty)\)

Step 1: Move all terms to one side

Start by moving all the terms to one side of the inequality to set it to zero. This helps us in factoring or applying the quadratic formula. The inequality becomes: \[10x^2 + 11x - 6 > 0\]

Step 2: Factor the quadratic expression

Next, attempt to factor the quadratic expression \(10x^2 + 11x - 6\) if possible. Look for two numbers that multiply to \(10 \times -6 = -60\) and add up to \(11\). These numbers are \(15\) and \(-4\). So, rewrite the middle term:\[10x^2 + 15x - 4x - 6 > 0\]Group and factor:\[5x(2x + 3) - 2(2x + 3) > 0\]Finally, factor by grouping:\[(5x - 2)(2x + 3) > 0\]

Step 3: Identify critical points

Set each factor equal to zero to find the critical points where the inequality may change. Solve for \(x\):1. \(5x - 2 = 0\) gives \(x = \frac{2}{5}\)2. \(2x + 3 = 0\) gives \(x = -\frac{3}{2}\) These critical points split the number line into intervals.

Step 4: Test intervals

Test the sign of the expression \((5x - 2)(2x + 3)\) in each interval defined by the critical points. Choose test points from each interval: 1. Interval \((-\infty, -\frac{3}{2})\), use \(x = -2\).2. Interval \((-\frac{3}{2}, \frac{2}{5})\), use \(x = 0\).3. Interval \((\frac{2}{5}, \infty)\), use \(x = 1\).Check the signs:- \(x = -2\): \((5(-2) - 2)(2(-2) + 3) = (-12)(-1) > 0\)- \(x = 0\): \((5(0) - 2)(2(0) + 3) = (-2)(3) < 0\)- \(x = 1\): \((5(1) - 2)(2(1) + 3) = (3)(5) > 0\)

Step 5: Write the solution in interval notation

Based on the sign of the product, the solution to the inequality \((5x - 2)(2x + 3) > 0\) includes intervals where the product is positive. These are:\((-\infty, -\frac{3}{2}) \cup (\frac{2}{5}, \infty)\)

Key Concepts

Factoring Quadratics

Factoring quadratics requires breaking down a quadratic expression into a product of simpler expressions. This process often involves identifying two numbers that both multiply to give the product of the quadratic's leading coefficient and constant term, while simultaneously adding up to the linear coefficient. For the inequality \(10x^2 + 11x - 6 > 0\), we need to find two numbers that multiply to \(-60\) (since \(10 \times -6 = -60\)) and add up to \(11\). These numbers are \(15\) and \(-4\). Next, we rewrite the middle term \(11x\) using these numbers:

• \(10x^2 + 15x - 4x - 6\)

Now, group and factor:

• \(5x(2x + 3) - 2(2x + 3)\)

Finally, factor by grouping the terms:

• \((5x - 2)(2x + 3) > 0\)

Factoring helps in identifying the critical points where the expression changes sign, which is essential for solving inequalities.

Critical Points in Inequalities

Critical points of certain inequalities like quadratic ones help us to understand where the inequality changes from true to false or vice versa. Once we have factored the quadratic expression, we set each factor equal to zero to find these critical points.For the inequality \((5x - 2)(2x + 3) > 0\), set each factor to zero:

• \(5x - 2 = 0\), which gives \(x = \frac{2}{5}\)
• \(2x + 3 = 0\), which gives \(x = -\frac{3}{2}\)

These critical points, \(x = \frac{2}{5}\) and \(x = -\frac{3}{2}\), indicate where the product changes. They divide the number line into intervals, making them key locations for determining the sign of the inequality in various parts of the number line.

Interval Notation

Interval notation provides a concise way to show the range of values that satisfy an inequality. Ordering these intervals helps us describe the solution set of our quadratic inequality.After establishing the critical points, we can divide the number line into intervals:

• \((-\infty, -\frac{3}{2})\)
• \((-\frac{3}{2}, \frac{2}{5})\)
• \((\frac{2}{5}, \infty)\)

The solution to our initial inequality, \((5x - 2)(2x + 3) > 0\), are portions where the product is positive. We use testing to determine the sign of the inequality in these intervals. Upon testing, the intervals with positive products are \((-\infty, -\frac{3}{2})\) and \((\frac{2}{5}, \infty)\).Interval notation then can express the solution as: \((-\infty, -\frac{3}{2}) \cup (\frac{2}{5}, \infty)\).

Sign Testing for Inequalities

Sign testing is crucial in identifying which intervals satisfy a quadratic inequality. After factoring and determining critical points, selecting test points within the determined intervals helps verify the inequality's status over those ranges.Consider the intervals \((-\infty, -\frac{3}{2})\), \((-\frac{3}{2}, \frac{2}{5})\), and \((\frac{2}{5}, \infty)\). By picking a point in each interval, you can evaluate the sign of the expression \((5x - 2)(2x + 3)\).- Test point \(x = -2\) in \((-\infty, -\frac{3}{2})\): - The result is positive, implying the inequality holds here.- Test point \(x = 0\) in \((-\frac{3}{2}, \frac{2}{5})\): - The result is negative, indicating the inequality doesn't hold.- Test point \(x = 1\) in \((\frac{2}{5}, \infty)\): - The result is positive, showing the inequality holds.These tests confirm that the product is positive in intervals \((-\infty, -\frac{3}{2})\) and \((\frac{2}{5}, \infty)\). Utilizing sign testing solidifies our solution range within the appropriate intervals.