Problem 26 - College Algebra with Modeling and Visualization

Question

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ -0.1 x^{2}+1=0.5 x $$

Step-by-Step Solution

Answer

The solutions are \( x = \frac{5 + \sqrt{65}}{-2} \) and \( x = \frac{5 - \sqrt{65}}{-2} \).

Step 1: Rewrite in Standard Form

To solve the quadratic equation, we first need to rewrite it in the standard form, which is \( ax^2 + bx + c = 0 \). Start with the equation: \(-0.1x^2 + 1 = 0.5x\). Move all terms to the left side to get: \(-0.1x^2 - 0.5x + 1 = 0\).

Step 2: Eliminate Decimals

To simplify our calculations, multiply the entire equation by 10 to eliminate decimals: \(-1x^2 - 5x + 10 = 0\). Now the quadratic equation is easier to work with: \(-x^2 - 5x + 10 = 0\).

Step 3: Solve the Quadratic Equation Using the Quadratic Formula

The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be used to find the roots of the equation. Here, \( a = -1 \), \( b = -5 \), and \( c = 10 \). Plug these values into the formula: \[ x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(-1)(10)}}{2(-1)} \].

Step 4: Calculate the Discriminant

The discriminant \( b^2 - 4ac \) is \((-5)^2 - 4(-1)(10) = 25 + 40 = 65\). The positive discriminant indicates two distinct real solutions.

Step 5: Compute the Solutions

Substitute the discriminant back into the quadratic formula: \[ x = \frac{5 \pm \sqrt{65}}{-2} \]. This gives us two solutions: \[ x = \frac{5 + \sqrt{65}}{-2} \] and \[ x = \frac{5 - \sqrt{65}}{-2} \]. Simplifying these fractions will provide the exact decimal values.

Step 6: Check the Solutions

Verify the solutions by substituting them back into the original equation \(-0.1x^2 + 1 = 0.5x\). This step ensures no arithmetic mistakes were made during calculations.

Key Concepts

Solving Quadratic Equations

Quadratic equations are mathematical expressions of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants and \( a eq 0 \). These equations describe parabolic curves, and finding their solutions means determining where these curves intersect the x-axis, if at all. To solve them, we can use different methods:

• Factoring: If the quadratic can be factored, you set each factor equal to zero to find the solutions.
• Completing the Square: This converts the quadratic into a perfect square trinomial, which can then be solved by taking the square root.
• Quadratic Formula: A universal method that can solve any quadratic equation, whether or not it can be easily factored.

More often than not, the quadratic formula is the go-to method because it's reliable regardless of the polynomial's complexity. Understanding these methods helps build a strong foundation in algebra and prepares you for future mathematical concepts.

Quadratic Formula

The quadratic formula is essential for solving quadratic equations and is particularly useful when the equation cannot be easily factored. The formula is given by \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. Each component of this formula has a role:

• -b: The opposite of \( b \), it's part of the equation that balances the signs.
• ± : This indicates two possible values, leading to the potential of two solutions for the equation.
• \( \sqrt{b^2 - 4ac} \): Known as the discriminant, this part under the square root determines the nature of the roots.

Plug the values of \( a \), \( b \), and \( c \) from the equation into the quadratic formula to find the solutions. This consistent method guarantees an answer, providing insight into whether we have real or complex solutions.

Discriminant

The discriminant is part of the quadratic formula, expressed as \( b^2 - 4ac \). It plays a critical role in determining the nature of the roots of the quadratic equation. Here are the potential scenarios:

• Positive discriminant: If \( b^2 - 4ac > 0 \), the equation has two distinct real roots.
• Zero discriminant: If \( b^2 - 4ac = 0 \), the equation has exactly one real root, or a double root.
• Negative discriminant: If \( b^2 - 4ac < 0 \), the equation has two complex roots.

Understanding the discriminant offers a quick assessment of what type of solutions to expect. In our example, where the discriminant is 65, we know there are two distinct real solutions.

Standard Form of a Quadratic Equation

Before solving any quadratic equation, it's usually helpful to convert it to the standard form, \( ax^2 + bx + c = 0 \). In this form:

• \( a \): The coefficient of \( x^2 \). It determines the "width" and direction (up or down) of the parabola.
• \( b \): The coefficient of \( x \), affecting the symmetry of the parabola.
• \( c \): The constant term, which moves the parabola up and down the y-axis.

Rewriting your equation in standard form makes it much easier to apply solving techniques, such as factoring or using the quadratic formula. For example, by rewriting \(-0.1x^2 + 1 = 0.5x\) as \(-x^2 - 5x + 10 = 0\), it was possible to straightforwardly identify \( a = -1 \), \( b = -5 \), and \( c = 10 \). This form also simplifies future calculations and ensures that all aspects of the equation are properly defined.