Problem 77 - Precalculus

Question

(2.5) If \(p(x)=2 x^{2}-x-3,\) in what intervals is \(p(x) \leq 0 ?\)

Step-by-Step Solution

Answer

The function is non-positive on the interval \([-1, \frac{3}{2}]\).

Step 1: Find the Roots of the Quadratic

First, identify the roots of the quadratic equation by solving \(2x^2 - x - 3 = 0\) using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \(a = 2\), \(b = -1\), and \(c = -3\). Calculate the discriminant, \(b^2 - 4ac = (-1)^2 - 4(2)(-3) = 1 + 24 = 25\). Since the discriminant is positive, the quadratic has two distinct real roots.

Step 2: Calculate the Roots

Use the quadratic formula to find the roots:\[x = \frac{-(-1) \pm \sqrt{25}}{2 \cdot 2}\]\[x = \frac{1 \pm 5}{4}\]This gives the roots \(x = \frac{6}{4} = \frac{3}{2}\) and \(x = \frac{-4}{4} = -1\).

Step 3: Determine the Sign in Each Interval

The quadratic \(p(x) = 2x^2 - x - 3\) can be written as \((x+1)(2x-3)\) using its roots. The roots divide the real number line into the intervals \((-\infty, -1)\), \((-1, \frac{3}{2})\), and \((\frac{3}{2}, \infty)\). Test a point from each interval to determine the sign of \(p(x)\):- For \(x = -2\) from \((-\infty, -1)\), \(p(-2) = 8 + 2 - 3 = 7 > 0\).- For \(x = 0\) from \((-1, \frac{3}{2})\), \(p(0) = -3 < 0\).- For \(x = 2\) from \((\frac{3}{2}, \infty)\), \(p(2) = 8 - 2 - 3 = 3 > 0\).

Step 4: Conclusion of Intervals

Based on the sign of \(p(x)\) in each interval, \(p(x) \leq 0\) is true in the interval \([-1, \frac{3}{2}]\). This includes the points \(x = -1\) and \(x = \frac{3}{2}\) since at these points, \(p(x) = 0\).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It provides a straightforward way to find the values of \( x \) where the equation equals zero.
The formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here's how it works:

• \( a \), \( b \), and \( c \) are constants in the quadratic equation.
• The expression under the square root, \( b^2 - 4ac \), is called the discriminant. It determines the nature of the roots.
• The \( \pm \) symbol indicates that there are generally two solutions: one for addition and one for subtraction.

The quadratic formula is not just a way of find solutions, but it also helps in understanding the characteristics of a quadratic equation, such as the number of real roots it possesses.

Roots of a Quadratic Equation

The roots of a quadratic equation are the values that make the equation equal to zero. These are also known as solutions or zeros of the quadratic equation.
In the exercise, the equation \( 2x^2 - x - 3 = 0 \) is solved using the quadratic formula to find its roots:

• The discriminant \( b^2 - 4ac = 25 \) is positive, hence the equation has two distinct real roots.
• By applying the quadratic formula, we find the roots to be \( x = \frac{3}{2} \) and \( x = -1 \).

These roots are critical because they help in determining where the quadratic function intersects the x-axis. Finding the roots allows one to understand the range in which the quadratic expression changes its sign.

Sign Analysis

Sign analysis is essential in solving quadratic inequalities like \( p(x) \leq 0 \). Once the roots are identified, they divide the real number line into intervals which can be tested to determine the sign of the expression in each section.
Here's how you can do it:

• Identify the intervals by using the roots \(-1\) and \(\frac{3}{2}\). The intervals here are \((-\infty, -1)\), \((-1, \frac{3}{2})\), and \((\frac{3}{2}, \infty)\).
• Select a test point from each interval and substitute it back into the function \( p(x) = 2x^2 - x - 3 \).

By determining the sign at these test points, we know:

• For \( x = -2 \), \( p(x) > 0 \), so the expression is positive in \((-\infty, -1)\).
• For \( x = 0 \), \( p(x) < 0 \), showing negativity in \((-1, \frac{3}{2})\).
• For \( x = 2 \), \( p(x) > 0 \), indicating positivity again in \((\frac{3}{2}, \infty)\).

The inequality \( p(x) \leq 0 \) holds true in the interval \([-1, \frac{3}{2}]\), where the expression's value is less than or equal to zero.