Problem 46 - Precalculus Student Solutions Manual 5th

Question

Solve each quadratic inequality. Write each solution set in interval notation. $$x(x+1)<12$$

Step-by-Step Solution

Answer

The solution set in interval notation is \( (-4, 3) \).

Step 1 - Move all terms to one side

First, move all terms to one side to set the inequality to zero. To do this, subtract 12 from both sides of the inequality: \[ x(x+1) - 12 < 0 \] which simplifies to: \[ x^2 + x - 12 < 0 \]

Step 2 - Factor the quadratic expression

Next, factor the quadratic expression to find its roots. The factors of \(x^2 + x - 12\) are \[ (x+4)(x-3) \]so the inequality can be rewritten as: \[ (x+4)(x-3) < 0 \]

Step 3 - Identify the critical points

Set each factor equal to zero to find the critical points:\[ x+4 = 0 \implies x = -4 \]\[ x-3 = 0 \implies x = 3 \]These points divide the number line into three intervals: \( (-\text{∞}, -4) \), \( (-4, 3) \), and \( (3, \text{∞}) \).

Step 4 - Test the intervals

Select a test point from each interval and substitute it into the inequality \( (x+4)(x-3) < 0 \) to determine where the inequality is satisfied:For \( (-\text{∞}, -4) \), use \(x = -5\): \[ (-5+4)(-5-3) = (-1)(-8) = 8 \] (not true)For \( (-4, 3) \), use \(x = 0\): \[ (0+4)(0-3) = 4(-3) = -12 \] (true)For \( (3, \text{∞}) \), use \(x = 4\): \[ (4+4)(4-3) = 8(1) = 8 \] (not true)

Step 5 - Write the solution in interval notation

Since the inequality \( (x+4)(x-3) < 0 \) is true only for the interval \( (-4, 3) \), the solution set in interval notation is: \[ (-4, 3) \]

Key Concepts

solving quadratic inequalities

To solve quadratic inequalities, such as our example \(x(x+1)<12\), we follow several steps to find the set of values that satisfy the inequality. When solving these types of inequalities:

• Move all terms to one side of the inequality to set the expression to zero.
• Factor the quadratic expression to identify key values, known as critical points.
• Divide the number line into intervals based on the critical points.
• Test points within each interval to check for truth in the inequality.
• Write the solution in interval notation to represent all valid values succinctly.

For our specific problem, we started by moving terms and simplifying the inequality to \(x^2 + x - 12 < 0\). By factoring, we transformed it into \((x+4)(x-3) < 0\), marking the key critical points.

factoring quadratic expressions

Factoring quadratic expressions is an essential step in solving quadratic inequalities. Given a standard quadratic form \(ax^2 + bx + c\), factoring involves rewriting it as a product of two binomials. For the expression \(x^2 + x - 12\), we need two numbers that multiply to \(-12\) (the constant term) and add up to \(1\) (the coefficient of \(x\)).

• These numbers are \(4-3=1\) and \(4* -3 = -12\).

So, the factors are \((x+4)(x-3)\).
Factoring lets us rewrite quadratic expressions, making it easier to identify critical points by setting each binomial to zero.

critical points

Critical points are values of \(x\) where the expression equals zero. For the inequality \((x+4)(x-3) < 0\):

• Setting \(x+4 = 0\) gives us \(x = -4\),
• Setting \(x-3 = 0\) results in \(x = 3\).

These points split the number line into three intervals: \(-∞, -4\), \(-4, 3\), and \(3, ∞\). By identifying critical points, we determine where to test values for the inequality.
Note: Critical points themselves won't necessarily satisfy the original inequality unless explicitly indicated.

interval notation

Interval notation is a simple way to express the set of solution values. It uses parentheses and brackets:

• \((a, b)\) represents all numbers between \(a\) and \(b\), not including \(a\) and \(b\) themselves.
• \([a, b)\) means to include \(a\) but not \(b\).
• For \((x+4)(x-3) < 0\), we found that the solution is true for the interval \((-4, 3)\),
• which means values between \(-4\) and \(3\) satisfy the inequality.

Interval notation is efficient, ensuring solutions are clear and concise.