Problem 19 - College Algebra with Modeling and Visualization

Question

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ \frac{1}{2} x^{2}-3 x+\frac{1}{2}=0 $$

Step-by-Step Solution

Answer

The solutions are \(x = 3 + 2\sqrt{2}\) and \(x = 3 - 2\sqrt{2}\).

Step 1: Identify the Coefficients

The given quadratic equation is \(\frac{1}{2}x^2 - 3x + \frac{1}{2} = 0\). Here, we identify the coefficients: \(a = \frac{1}{2}\), \(b = -3\), and \(c = \frac{1}{2}\).

Step 2: Use the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Substitute the identified coefficients into the formula.

Step 3: Calculate the Discriminant

Calculate the discriminant using the formula \(b^2 - 4ac\). Here, the discriminant is \((-3)^2 - 4 \times \frac{1}{2} \times \frac{1}{2}\). Simplifying gives \(9 - 1 = 8\).

Step 4: Substitute into the Quadratic Formula

Now substitute \(b = -3\), \(\sqrt{8}\), and \(a = \frac{1}{2}\) into the quadratic formula: \(x = \frac{3 \pm \sqrt{8}}{1}\).

Step 5: Simplify the Solutions

Simplify \(\sqrt{8}\) as \(2\sqrt{2}\) and solve: \(x = 3 \pm 2\sqrt{2}\). This results in two solutions \(x = 3 + 2\sqrt{2}\) and \(x = 3 - 2\sqrt{2}\).

Step 6: Check the Solutions

Substitute \(x = 3 + 2\sqrt{2}\) back into the original equation and verify it equals zero. Repeat for \(x = 3 - 2\sqrt{2}\). Both should satisfy the original equation.

Key Concepts

Quadratic Formula

In algebra, solving quadratic equations is a fundamental skill that you often use. These equations are typically written in the form

• ax² + bx + c = 0

where 'a,' 'b,' and 'c' are constants. The quadratic formula is a magical tool that helps you solve any quadratic equation. It is given by

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

This formula allows you to find the values of x that satisfy the equation, assuming you have the correct coefficients for a, b, and c.
To use the formula:

• First, identify the coefficients in your quadratic equation, just as we did in the original exercise.
• Plug these coefficients into the formula.

It's a straightforward process that always follows these steps. This is particularly useful because it works even when the quadratic isn't factorable. Remember to first calculate the discriminant, which we'll talk more about in the next section.

Discriminant

The discriminant is a special term in the quadratic formula located under the square root:

• \(b^2 - 4ac\)

This small expression plays a big role in determining the nature of the solutions for a quadratic equation.
The value of the discriminant tells us:

• If it’s positive, the equation has two distinct real roots.
• If it’s zero, the equation has exactly one real root, or a repeated root.
• If it’s negative, there are no real solutions but two complex solutions.

In our specific example, the discriminant was calculated as

• \(8\)

(as shown in Step 3 of the original solution), which is positive. This tells us right away that our equation has two distinct real solutions. Calculating the discriminant is quick and gives you a clear picture of what kind of solutions to expect.

Simplifying Expressions

Once you have plugged values into the quadratic formula and evaluated the expression, you're left with a solution that often needs simplifying. Simplifying expressions helps in making the solutions more understandable and neat.
In the original exercise, we simplified:

• \(\sqrt{8}\)

This can be simplified further into

• \(2\sqrt{2}\)

This simplification makes the expression cleaner and easier to work with. Sometimes, simplification reveals further insights into relationships or properties within the problem.
Remember to always check your work by plugging the simplified solutions back into the original equation. This ensures that no errors were made during the simplification process. It might be a bit tedious but it’s a great way to confirm the solution.