Problem 42 - Precalculus Student Solutions Manual 5th

Question

Solve each quadratic inequality. Write each solution set in interval notation. $$3 x^{2}+x \leq 4$$

Step-by-Step Solution

Answer

The solution set is \(\left[ -\frac{4}{3}, 1 \right]\).

Step 1: Write the inequality in standard form

First, rewrite the given inequality in the standard form of a quadratic inequality. The standard form is \[ ax^2 + bx + c \triangleq 0 \] Move all terms to one side of the inequality:\[ 3x^2 + x - 4 \triangleq 0 \]

Step 2: Solve the quadratic equation

To find the critical points, solve the corresponding quadratic equation \[ 3x^2 + x - 4 = 0 \] Use the quadratic formula \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] In this inequality, the coefficients are: \[ a = 3, \, b = 1, \, c = -4 \]

Step 3: Calculate the discriminant

Calculate the discriminant \[ \triangle = b^2 - 4ac \] Substitute the coefficients: \[ \triangle = 1^2 - 4 \cdot 3 \cdot (-4) = 1 + 48 = 49 \]

Step 4: Find the roots using the quadratic formula

Substitute the values into the quadratic formula: \[ x = \frac{-1 \pm \sqrt{49}}{2 \cdot 3} \] Simplify to get the roots: \[ x = \frac{-1 \pm 7}{6} \] This gives: \[ x = 1 \text{ or } x = -\frac{4}{3} \]

Step 5: Determine sign of the quadratic expression

Using the roots obtained, test the intervals \[ (-\infty, -\frac{4}{3}), \left[-\frac{4}{3}, 1\right], (1, \infty) \] Test points from each interval in the original inequality. Typically, test points such as \( x = -2 \), \( x = 0 \), and \( x = 2 \):- For \( x = -2 \): \[ 3(-2)^2 + (-2) - 4 = 12 - 2 - 4 = 6 \] (not less than or equal to 0)- For \( x = 0 \): \[ 3(0)^2 + 0 - 4 = -4 \] (less than or equal to 0)- For \( x = 2 \): \[ 3(2)^2 + 2 - 4 = 12 + 2 - 4 = 10 \] (not less than or equal to 0)

Step 6: Write the solution in interval notation

Combine the intervals where the quadratic expression is less than or equal to zero. The solution set for \[ 3x^2 + x - 4 \triangleq 0 \] is \[ \left[ -\frac{4}{3}, 1 \right] \]

Key Concepts

solving quadratic inequalities

Solving quadratic inequalities involves finding the range of values for the variable that satisfy the inequality condition. Let's break this down step by step to make it clearer.

• First, you need to rewrite the inequality in standard form. This means getting all terms on one side, so you have something like \( ax^2 + bx + c \triangleq 0 \) or \( ax^2 + bx + c \triangleq k \).
• Next, solve the corresponding quadratic equation. This will give you the critical points or the roots of the equation. In our case, these were \( x = 1 \) and \( x = -\frac{4}{3} \).
• Then, identify the intervals between and around these roots: \( (-\infty, -\frac{4}{3}) \), \( \left[-\frac{4}{3}, 1\right] \), and \( (1, \infty) \).
• Test a value from each of these intervals in the original inequality to see if it satisfies the condition.

Once you've tested the points, you combine the intervals where the conditions are satisfied. This gives you the solution set.

interval notation

Once you've determined the intervals that satisfy the quadratic inequality, you need to express the solution using interval notation. Interval notation is a way of writing subsets of the real number line:

• Use brackets \( [ \) and \( ] \) to denote inclusive intervals, meaning the endpoints are included in the interval.
• Use parentheses \( ( \) and \( ) \) to denote exclusive intervals, meaning the endpoints are not included.

For example, the solution to our problem, \( \left[ -\frac{4}{3}, 1 \right] \), shows that all values between \( -\frac{4}{3} \) and \( 1 \), including \( -\frac{4}{3} \) and \( 1 \) themselves, satisfy the inequality. Interval notation is efficient for writing the range of values quickly and clearly.

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations. It is used when a quadratic equation is in the form \( ax^2 + bx + c = 0 \). The formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here’s how it works step-by-step:

• Identify the coefficients \( a \), \( b \), and \( c \).
• Calculate the discriminant, \( \Delta = b^2 - 4ac \).
• Substitute these values into the formula to find the values of \( x \).

The solution might give you two real roots, one real root, or complex roots, depending on the value of the discriminant. In our example, using the coefficients \( a = 3 \), \( b = 1 \), and \( c = -4 \), we got \( x = 1 \) and \( x = -\frac{4}{3} \) as the roots by applying the quadratic formula.

discriminant

The discriminant is an important part of the quadratic formula. It helps determine the nature and number of the roots of a quadratic equation. The discriminant is given by the formula:

\[ \Delta = b^2 - 4ac \]

Depending on the value of \( \Delta \):

• If \( \Delta > 0 \), the quadratic equation has two distinct real roots.
• If \( \Delta = 0 \), the quadratic equation has exactly one real root (a repeated root).
• If \( \Delta < 0 \), the quadratic equation has two complex roots.

In our example, we calculated the discriminant as \( \Delta = 49 \). Since \( \Delta > 0 \), this told us that the quadratic equation \( 3x^2 + x - 4 = 0 \) has two distinct real roots, which are \( x = 1 \) and \( x = -\frac{4}{3} \). Understanding the discriminant is crucial for determining how the function behaves and solving inequalities.