Problem 58 - Algebra for College Students

Question

Expressing solutions to the nearest one-thousandth. $$x^{2}+8 x+2=0$$

Step-by-Step Solution

Answer

The solutions, rounded to the nearest one-thousandth, are \( x_1 = -0.259 \) and \( x_2 = -7.742 \).

Step 1: Identify the quadratic equation

We start with the given quadratic equation: \[ x^2 + 8x + 2 = 0 \] This is a standard form of a quadratic equation \( ax^2 + bx + c = 0 \) with \( a = 1 \), \( b = 8 \), and \( c = 2 \).

Step 2: Apply the quadratic formula

To solve the quadratic equation, we use the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Substitute \( a = 1 \), \( b = 8 \), and \( c = 2 \) into the formula.

Step 3: Calculate the discriminant

Calculate \( b^2 - 4ac \):\[ b^2 - 4ac = 8^2 - 4 \times 1 \times 2 = 64 - 8 = 56 \]Since 56 is positive, there are two real solutions.

Step 4: Compute the square root of the discriminant

Next, find \( \sqrt{56} \):\[ \sqrt{56} \approx 7.483 \] We'll use this approximation in the next step to calculate the roots.

Step 5: Find the solutions using the quadratic formula

Substitute \( b = 8 \) and \( \sqrt{56} \approx 7.483 \) into the quadratic formula:\[ x = \frac{-8 \pm 7.483}{2} \]Compute the two solutions:1. \[ x_1 = \frac{-8 + 7.483}{2} = \frac{-0.517}{2} \approx -0.259 \]2. \[ x_2 = \frac{-8 - 7.483}{2} = \frac{-15.483}{2} \approx -7.742 \]

Step 6: Round to the nearest one-thousandth

Ensure to round each solution to the nearest one-thousandth:\( x_1 = -0.259 \)\( x_2 = -7.742 \)

Key Concepts

The Quadratic Formula

To tackle quadratic equations, the quadratic formula is your go-to tool. This formula works for any quadratic equation of the form \( ax^2 + bx + c = 0 \). The coefficients \( a \), \( b \), and \( c \) are numbers from the equation. The quadratic formula is given by:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
This powerful formula can give you the solutions, or roots, of the equation. The "\( \pm \)" symbol indicates there are two possible solutions, one found by adding the square root part, and another by subtracting it. You'll see how these solutions vary depending on the different values of the discriminant, which is the expression under the square root, \( b^2 - 4ac \).

Understanding the Discriminant

The discriminant in the quadratic formula, \( b^2 - 4ac \), holds the key to understanding the nature of the solutions. Depending on its value, the equation may have different types of solutions:

• If the discriminant is positive, like our example of 56, there are two distinct real solutions.
• If it's zero, you'll find exactly one real solution (also called a repeated root).
• If negative, there are no real solutions, but rather two complex solutions.

Recognizing the role of the discriminant helps in anticipating the type of solution to expect. It's a quick check before solving that can save time and adjust your approach efficiently.

Real Solutions of a Quadratic Equation

Real solutions occur when the discriminant is zero or positive. For our example, with a discriminant of 56, two real solutions exist. By solving the quadratic formula, you get values for \( x \) that make the equation true. These solutions were calculated as follows:
- \( x_1 = \frac{-8 + 7.483}{2} = -0.259 \)
- \( x_2 = \frac{-8 - 7.483}{2} = -7.742 \)
These represent real numbers, meaning they can be plotted on a number line. Real solutions indicate points where the parabola, the graph of the quadratic equation, crosses the x-axis. In practical terms, it's where the equation "comes alive" in the realm of real numbers.

Rounding Decimals to the Nearest Thousandth

When dealing with real-number solutions, it’s essential that you express them with the necessary precision, often to reflect practical scenarios or specific instructions. In our case, each solution is rounded to the nearest thousandth. Remember these steps to get it right:

• Look at the fourth decimal place (if present), which determines if the rounding applies upwards or stays the same.
• If the fourth decimal is 5 or more, round the third decimal up by one.
• If it's less than 5, keep the third decimal unchanged.

For instance, the calculated solution of \(-0.259\) is rounded directly as there are no further digits. Similarly, \(-7.742\) follows straightforwardly, giving a neat and tidy answer suitable for presentation or further calculation.