Problem 41 - Algebra 1

Question

Use the quadratic formula to solve the equation. $$-\frac{1}{2} x^{2}+6 x+13=0$$

Step-by-Step Solution

Answer

The solutions for the equation are \(x_1 = 2 - \sqrt{15}\; and \;x_2 = 2 + \sqrt{15}\)

Step 1: Identify the coefficients

From the equation \(-\frac{1}{2} x^{2}+6 x+13=0\), we can identify the coefficients as \(a = -\frac{1}{2}\), \(b = 6\), and \(c = 13\).

Step 2: Substitute these coefficients into the quadratic formula

Replace \(a\), \(b\), and \(c\) in the formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(-\frac{1}{2}\), \(6\), and \(13\) respectively, we get \(x = \frac{-6 \pm \sqrt{6^2 - 4(-\frac{1}{2})*13}}{2*(-\frac{1}{2})}\).

Step 3: Simplify the equation

Simplifying the above expression gives us two potential solutions for \(x\): \(x = \frac{-6 + \sqrt{36 + 26}}{-1}\) and \(x = \frac{-6 - \sqrt{36 + 26}}{-1}\)

Step 4: Calculate the solutions

These are the calculated solutions: \(x_1 = 2 - \sqrt{15}\; and \;x_2 = 2 + \sqrt{15}\)

Key Concepts

quadratic equations

Quadratic equations are polynomial equations of degree two. This means they take the general form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are coefficients and \( x \) represents the variable being solved for. Quadratic equations are prevalent in various fields such as physics, engineering, and finance because they describe parabolic relationships.

To solve quadratic equations, several methods may be used, including factoring, completing the square, or using the quadratic formula. The quadratic formula is applicable to all quadratic equations and is often used when factoring is not easily achievable.

• Square term (\( ax^2 \)): The variable \( x \) is multiplied by itself and then by the coefficient \( a \).
• Linear term (\( bx \)): The variable \( x \) is multiplied by the coefficient \( b \).
• Constant term (\( c \)): A standalone number without the variable \( x \).

coefficients

In a quadratic equation, coefficients are critical as they determine the shape and position of the parabola related to the equation. For example, in the quadratic equation \(-\frac{1}{2}x^2 + 6x + 13 = 0\):

• \( a = -\frac{1}{2} \): This coefficient influences the opening direction and the width of the parabola. A negative \( a \) value means the parabola opens downwards.
• \( b = 6 \): This coefficient affects the slope of the tangent lines to the parabola at its intersection points.
• \( c = 13 \): This is the constant or y-intercept value, representing where the parabola crosses the y-axis.

Understanding these coefficients allows one to graph the equation confidently and solve it using methods like the quadratic formula. When you substitute values for \( a \), \( b \), and \( c \) into the quadratic formula, it becomes straightforward to find the solutions of the equation.

solving equations

Solving quadratic equations often involves using the quadratic formula. This formula derives from the process of completing the square and provides a direct way to find the roots of a quadratic equation. For a given equation \( ax^2 + bx + c = 0 \), the formula is:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here's how this works in practice:

• Identify \( a \), \( b \), and \( c \) from the equation, e.g., \(-\frac{1}{2}, 6, \text{ and } 13\).
• Plug these values into the quadratic formula.
• Calculate the discriminant, \( b^2 - 4ac \). A positive discriminant indicates two real solutions, a zero discriminant one real solution, and a negative discriminant implies complex solutions.
• Solve for \( x \) by completing the arithmetic as instructed by the formula.

This structured approach simplifies the problem-solving process, ensuring accuracy and efficiency. In the example provided, this method yields solutions that involve both addition and subtraction through the square root, demonstrating the utility of the quadratic formula.