Problem 38 - Precalculus: Functions and Graphs

Question

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$2 x^{2}+5 x-12=0$$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{3}{2}\) and \(x = -4\).

Step 1: Identify the type of problem

We need to solve the quadratic equation \(2x^2 + 5x - 12 = 0\). This problem involves finding the values of \(x\) that make the equation true.

Step 2: Use the Quadratic Formula

For a quadratic equation in the form \(ax^2 + bx + c = 0\), the solutions can be found using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, \(a = 2\), \(b = 5\), and \(c = -12\).

Step 3: Calculate the Discriminant

The discriminant \(b^2 - 4ac\) determines the nature of the roots. Substitute \(a = 2\), \(b = 5\), and \(c = -12\) into the formula to get \(5^2 - 4 \times 2 \times (-12) = 25 + 96 = 121\). Since 121 is a positive perfect square, we will have two real and distinct solutions.

Step 4: Solve for \(x\)

Substitute the values of \(a\), \(b\), \(c\), and the square root of the discriminant into the quadratic formula:\[x = \frac{-5 \pm \sqrt{121}}{4}\]This simplifies to:\[x = \frac{-5 \pm 11}{4}\]

Step 5: Simplify each solution

This results in two values for \(x\):1. \[x = \frac{-5 + 11}{4} = \frac{6}{4} = \frac{3}{2}\]2. \[x = \frac{-5 - 11}{4} = \frac{-16}{4} = -4\]

Step 6: Express the solution as an interval

The solutions \(x = \frac{3}{2}\) and \(x = -4\) are points, not intervals, as this equation involves finding specific values of \(x\) for a quadratic equation.

Key Concepts

quadratic formula

The quadratic formula is a crucial tool for solving quadratic equations like \( ax^2 + bx + c = 0 \). The formula is given as:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

Using this formula, we can find the values of \( x \) that satisfy the equation. The variables \( a \), \( b \), and \( c \) are the coefficients from the quadratic equation. Identifying these values is the first step. In our example, \( a = 2 \), \( b = 5 \), and \( c = -12 \).
Inserting these into the formula allows us to calculate potential roots for the equation. It's essential to perform substitutions correctly and work through the operations one step at a time to avoid errors.

discriminant

The discriminant is the part of the quadratic formula under the radical sign: \( b^2 - 4ac \). This value reveals important information about the nature of the roots for the quadratic equation.

• If the discriminant is greater than 0, the equation has two real and distinct solutions.
• If it's equal to 0, there is one real solution or a repeated root.
• If it's less than 0, the equation has no real solutions, but two complex ones.

In our case, substituting \( a = 2 \), \( b = 5 \), and \( c = -12 \) into \( b^2 - 4ac \) gives \( 25 + 96 = 121 \). Since 121 is a positive perfect square, it confirms that the quadratic equation has two real and distinct solutions.

real and distinct solutions

When solving the quadratic equation, we found the discriminant to be positive, which meant the solutions are real and distinct. Let's understand why this happens.
The discriminant being positive indicates that the quadratic graph crosses the x-axis at two different points. That's why the equation has two separate values for \( x \).
Using the quadratic formula, after inserting our specific values, the solutions worked out to be:

• \( x = \frac{-5 + 11}{4} = \frac{3}{2} \)
• \( x = \frac{-5 - 11}{4} = -4 \)

These results demonstrate both solutions are different numbers, further confirming that they are distinct.

interval notation

Interval notation is a way of representing solutions, particularly for inequalities, using intervals on the real number line. However, for quadratic equations like ours, which yield specific solutions rather than intervals, we use individual points.
In the context of this exercise, the solutions \( x = \frac{3}{2} \) and \( x = -4 \) are specific points. They can't be expressed as intervals because they aren't ranges of values. These points indicate where the quadratic function's graph intersects the x-axis.
It's important to realize that although interval notation is a helpful tool to communicate ranges, it's not always applicable in cases where we find specific roots or zeros of an equation.