Problem 47 - Algebra 1: Concepts and Skills

Question

Use the quadratic formula to solve the equation. If the solution involves radicals, round to the nearest hundredth. $$-\frac{1}{2} x^{2}+6 x+13=0$$

Step-by-Step Solution

Answer

The solutions to the equation are \(x_1 = -1.87\) and \(x_2 = 13.87\).

Step 1: Identify a, b, and c

The general form of a quadratic equation is \(ax^2 + bx + c = 0\). In the given equation, \(a = -1/2\), \(b = 6\), and \(c = 13\).

Step 2: Substitute into Quadratic Formula

Place the values of a, b, c in the quadratic equation formula. \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-6 \pm \sqrt{{(6)^2 - 4*(-1/2)*13}}}}{2*(-1/2)} = \frac{{-6 \pm \sqrt{{36 + 26}}}}{-1} = \frac{{-6 \pm \sqrt{{62}}}}{-1}\).

Step 3: Calculate the Results

Calculate the values of x. The equation splits into two based on plus or minus in the square root. The two solutions are \(x_1 = \frac{{-6 + \sqrt{{62}}}}{-1} = 6 - \sqrt{{62}}\) and \(x_2 = \frac{{-6 - \sqrt{{62}}}}{-1} = 6 + \sqrt{{62}}\). As per the instruction, let's round these solutions to the nearest hundredths and evaluate to get \(x_1 = -1.87\) and \(x_2 = 13.87\).

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree two, typically written in the standard form:

• \( ax^2 + bx + c = 0 \)

where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. An important aspect of quadratic equations is that they can have two solutions, one solution, or no real solutions depending on the discriminant \( b^2 - 4ac \).

The term "discriminant" might sound complex, but it's simply the expression under the square root in the quadratic formula. If the discriminant is positive, there are two real solutions. If it's zero, there's exactly one real solution, and if negative, the solutions are complex (non-real).

In our specific problem, we have an equation: \(-\frac{1}{2} x^{2}+6 x+13=0\), where \( a = -1/2 \), \( b = 6 \), and \( c = 13 \). By using the quadratic formula, we address each component and calculate the solutions.

Solutions with Radicals

Radicals are expressions that include a square root, cube root, or higher roots. When solving quadratic equations, we often encounter a radical within the quadratic formula:

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

In this formula, the expression under the square root \( b^2 - 4ac \) can be positive, zero, or negative, affecting the type and nature of the solutions.

In our exercise, substituting \( a = -1/2 \), \( b = 6 \), and \( c = 13 \), we found that the expression under the square root simplifies to \( \sqrt{62} \). This is a radical term because \( 62 \) is not a perfect square and thus \( \sqrt{62} \) cannot be simplified to an exact numerical value without using a calculator.

We then consider two scenarios from the quadratic formula: one where we add the square root and one where we subtract it, providing two potential solutions. The solutions in radical form were determined as \( x_1 = 6 - \sqrt{62} \) and \( x_2 = 6 + \sqrt{62} \).

Rounding to Nearest Hundredth

Rounding is a process used to simplify numbers to a desired decimal place, making them easier to work with or read. In this exercise, after finding solutions with radicals, it was necessary to round them to the nearest hundredth for clarity and precision.

Each solution \( x_1 \) and \( x_2 \) derived from the quadratic formula involves evaluating the expression containing \( \sqrt{62} \):

• For \( x_1 \), calculate \( 6 - \sqrt{62} \)
• For \( x_2 \), calculate \( 6 + \sqrt{62} \)

After using a calculator, we find the decimal approximations:

• \( x_1 \approx -1.87 \)
• \( x_2 \approx 13.87 \)

Rounding these numbers to the nearest hundredth ensures that we present the solutions in a practical, usable form. This practice is common, especially when dealing with roots and irrational numbers, to maintain both accuracy and simplicity.