Problem 39 - Precalculus: Functions and Graphs

Question

Solve the equation or inequality. Express the solutions in terms of intervals whenever possible. $$x(3 x+4)=5$$

Step-by-Step Solution

Answer

Solutions are \(x = \frac{-2 + \sqrt{19}}{3}\) and \(x = \frac{-2 - \sqrt{19}}{3}\).

Step 1: Expand and Rearrange

First, expand the left side of the equation. We have: \[x(3x + 4) = 3x^2 + 4x\]So the equation becomes:\[3x^2 + 4x = 5\]Rearrange the equation to set it to zero form:\[3x^2 + 4x - 5 = 0\]

Step 2: Identify Coefficients

Identify the coefficients in the quadratic equation:\[3x^2 + 4x - 5 = 0\]Here, the coefficients are: \ a = 3, \ b = 4, \ c = -5.

Step 3: Use the Quadratic Formula

Apply the quadratic formula to solve for \(x\):\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}\]Substitute the coefficients \(a = 3\), \(b = 4\), and \(c = -5\) into the formula:\[x = \frac{{-4 \pm \sqrt{{4^2 - 4 \times 3 \times (-5)}}}}{2 \times 3}\]

Step 4: Calculate the Discriminant

Compute the discriminant \(b^2 - 4ac\):\[16 - 4 \times 3 \times (-5) = 16 + 60 = 76\]The discriminant is 76.

Step 5: Solve for x

Substitute the value of the discriminant back into the quadratic formula:\[x = \frac{{-4 \pm \sqrt{76}}}{6}\]Simplify \(\sqrt{76}\):\[\sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19}\]Thus:\[x = \frac{{-4 \pm 2\sqrt{19}}}{6}\]

Step 6: Simplify the Solutions

Split the expression into two solutions and simplify:\[x = \frac{{-4 + 2\sqrt{19}}}{6} \quad \text{or} \quad x = \frac{{-4 - 2\sqrt{19}}}{6}\]Upon simplifying both:\[x = \frac{{-2 + \sqrt{19}}}{3} \quad \text{or} \quad x = \frac{{-2 - \sqrt{19}}}{3}\]

Step 7: Express Solutions As Intervals

Since the solutions involve square roots, try to evaluate them to express as intervals. However, these particular solutions will be better left in this simplified form. Neither represents an interval occurring on the number line without further context.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool that helps you solve any quadratic equation of the form \(ax^2 + bx + c = 0\). This is particularly useful because not all quadratic equations can be easily factored. The formula is:

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

This formula gives you the solution for \(x\) directly by substituting the coefficients \(a\), \(b\), and \(c\) from the quadratic equation. Using this formula, you can handle complex roots and ensure all mathematical precision is maintained.
The process starts by identifying the coefficients \(a\), \(b\), and \(c\) from the equation, then substituting them into the formula. It’s essential to handle calculations carefully, especially under the square root. This portion of the formula is called the discriminant, which we will explore further in the next section.

Discriminant

The discriminant is a key piece of the quadratic formula, given by \(b^2 - 4ac\). Its value tells us important information about the nature of the roots of the quadratic equation:

• If the discriminant is positive, the quadratic equation has two distinct real roots.
• If the discriminant is zero, it has exactly one real root (or a repeated real root).
• If the discriminant is negative, there are no real roots, but instead, two complex roots.

In the original exercise, the discriminant calculates as \(16 + 60 = 76\). Since 76 is positive, this confirms that the equation has two distinct real roots. Calculating the discriminant forms an integral step because it informs whether further algebraic manipulations are needed and what type of number solutions to expect.

Interval Notation

Interval notation is a method of writing down a set of numbers, typically used to express solutions to inequalities. Unlike equations, inequalities often have ranges or intervals as solutions. For example, if a solution set consists of all \(x\) such that

• 0 \(< x < 5\),

we can express it as \((0, 5)\) in interval notation. Here:

• Round brackets \(()\) are used to denote that endpoints are not included in the interval.
• Square brackets \([]\) are used if you want to include the endpoints.

However, the original problem's solutions, \(x = \frac{{-2 + \sqrt{19}}}{3}\) and \(x = \frac{{-2 - \sqrt{19}}}{3}\), do not lend themselves directly to interval notation since they're individual solutions rather than ranges. In such cases, interval notation is typically not used unless translating into contextualized ranges.