Problem 37 - Precalculus

Question

Use the quadratic formula to solve each equation. In Exercises \(34-39,\) give two forms for each solution: an expression containing a radical and a calculator approximation rounded off to two decimal places. $$x(3 x+8)=-2$$

Step-by-Step Solution

Answer

Solutions: \(-\frac{4 \pm \sqrt{10}}{3}\), approx. \(-0.46, -1.54\).

Step 1: Expand and Rearrange the Equation

Start by expanding the equation. Multiply \(x\) by \((3x + 8)\) to obtain: \[x(3x + 8) = 3x^2 + 8x.\]Rearrange the equation to move \(-2\) to the left side, resulting in:\[3x^2 + 8x + 2 = 0.\]

Step 2: Identify the Coefficients

Identify the coefficients needed for the quadratic formula. For the equation \(3x^2 + 8x + 2 = 0\), the coefficients are:- \(a = 3\)- \(b = 8\)- \(c = 2\).

Step 3: Apply the Quadratic Formula

The quadratic formula is given by:\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}.\]Substitute the coefficients into the formula:\[x = \frac{{-8 \pm \sqrt{{8^2 - 4 \cdot 3 \cdot 2}}}}{2 \cdot 3}.\]

Step 4: Simplify Inside the Square Root

Calculate the discriminant \(b^2 - 4ac\):\[8^2 - 4 \cdot 3 \cdot 2 = 64 - 24 = 40.\]So the expression inside the square root becomes \(\sqrt{40}\).

Step 5: Simplify \(\sqrt{40}\)

Write \(\sqrt{40}\) in its simplest radical form:\[\sqrt{40} = \sqrt{4 \times 10} = \sqrt{4} \times \sqrt{10} = 2\sqrt{10}.\]

Step 6: Write the Solution without Approximation

Substitute \(\sqrt{40}\) as \(2\sqrt{10}\) back into the quadratic formula:\[x = \frac{{-8 \pm 2\sqrt{10}}}{6}.\]Simplify further:\[x = \frac{{-8 \pm 2\sqrt{10}}}{6} = \frac{{-4 \pm \sqrt{10}}}{3}.\]

Step 7: Calculator Approximation

Calculate the approximate values of the roots using a calculator. Evaluate both:1. \(x = \frac{{-4 + \sqrt{10}}}{3} \approx -0.46\).2. \(x = \frac{{-4 - \sqrt{10}}}{3} \approx -1.54\).

Key Concepts

Solutions of Quadratic Equations

When approaching quadratic equations, we often aim to find values of \(x\) that satisfy the equation, known as solutions or roots. In general, quadratic equations take the form \(ax^2 + bx + c = 0\). To solve them, we can use several methods, one of which is the quadratic formula given by:\[x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a}.\]This formula is derived from the process of completing the square, and it gives the solutions of any quadratic equation. Here's a simple breakdown of the steps involved:

• Identify the coefficients \(a, b,\) and \(c\) from the equation.
• Plug these values into the quadratic formula.
• The expression \(\pm\) indicates that there will generally be two solutions: one involving addition and the other subtraction.
• These solutions can be expressed in radical form or decimal form.

This approach is universal and works even when the other methods such as factoring are not applicable. Understanding this formula gives you a reliable tool for finding solutions to any quadratic equation.

Radical Expressions

A radical expression involves a root, typically a square root in the context of quadratic formula usage. Consider the expression under the square root in the quadratic formula: \(b^2 - 4ac\). This part is called the discriminant, and it dictates the nature of the solutions.

The Discriminant and Its Impact

- If \(b^2 - 4ac > 0\), the equation has two distinct real solutions
- If \(b^2 - 4ac = 0\), the equation has exactly one real solution, often called a double root
- If \(b^2 - 4ac < 0\), the equation has no real solutions, as the roots are complexFor our particular solution, \(b^2 - 4ac = 40\), which is positive, affirming two real roots. When simplifying the square root of a non-perfect square like \(\sqrt{40}\), factor it into smaller components such as \(\sqrt{4 \times 10}\) to obtain \(2\sqrt{10}\). This not only simplifies the expression but also makes manual calculations easier.Deeper comprehension of radical expressions aids in recognizing patterns and simplifying other similar problems in mathematics.

Calculator Approximations

Once you have the solutions in their simplified radical form, it often becomes necessary to obtain decimal approximations, especially for practical applications. A calculator is used to evaluate the precise numerical value.

Process of Calculator Approximations

- First, ensure the expression is simplified as much as possible, such as transforming \(\sqrt{40}\) to \(2\sqrt{10}\).- Input the simplified expression into a calculator to derive approximate decimal values.In our case, after substituting and simplifying as \(\frac{{-4 \pm \sqrt{10}}}{3}\), the calculations provide two approximate values:

• \(x = 0.46\)
• \(x = -1.54\)

Approximation is valuable in real-world applications where exact values are usable or intuitive for presenting or reporting results. Remember, calculations depend on the precision of the device used, so retrace your steps to ensure accuracy.