Problem 11 - College Algebra with Modeling and Visualization

Question

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ 6 x^{2}+\frac{5}{2}=8 x $$

Step-by-Step Solution

Answer

The solutions are \(x = \frac{5}{6}\) and \(x = \frac{1}{2}\).

Step 1: Rearrange the Equation

Start by moving all terms to one side of the equation to set it to zero. The given equation is \(6x^2 + \frac{5}{2} = 8x\). Subtract \(8x\) from both sides to get: \(6x^2 - 8x + \frac{5}{2} = 0\).

Step 2: Clear the Fraction

To eliminate the fraction, multiply every term in the equation by 2. This gives us: \(12x^2 - 16x + 5 = 0\).

Step 3: Use the Quadratic Formula

The equation is in the form \(ax^2 + bx + c = 0\) with \(a = 12\), \(b = -16\), and \(c = 5\). Use the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 4: Calculate the Discriminant

Calculate the discriminant \(b^2 - 4ac\). Substitute the values: \((-16)^2 - 4\cdot12\cdot5 = 256 - 240 = 16\).

Step 5: Solve for x Using the Quadratic Formula

Plug the values into the quadratic formula: \(x = \frac{16 \pm \sqrt{16}}{24}\). Simplify under the square root: \(\sqrt{16} = 4\). This results in \(x = \frac{16 \pm 4}{24}\).

Step 6: Find the Solutions

Solve for the two possible solutions: \(x_1 = \frac{16 + 4}{24} = \frac{20}{24} = \frac{5}{6}\) and \(x_2 = \frac{16 - 4}{24} = \frac{12}{24} = \frac{1}{2}\).

Step 7: Check the Solutions

Substitute \(x_1 = \frac{5}{6}\) into the original equation: \(6\left(\frac{5}{6}\right)^2 + \frac{5}{2} = 8\left(\frac{5}{6}\right)\). Verify both sides equal. Do the same for \(x_2 = \frac{1}{2}\). Both solutions satisfy the original equation.

Key Concepts

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). It allows you to find the values of \(x\) that satisfy the equation by using a specific formula:

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

Here, \(a\), \(b\), and \(c\) are coefficients from the quadratic equation, and \(x\) represents the unknowns we want to solve for. This formula works for any quadratic equation and is especially useful when the equation does not factor easily.

• \(-b\) refers to the opposite of \(b\), so always change the sign of \(b\).
• The \(±\) symbol means you will calculate two solutions: one with plus and one with minus.

The quadratic formula not only helps you find the solutions quickly but also gives you insight into the nature of these solutions through the discriminant, \(b^2 - 4ac\).

Discriminant

The discriminant is a part of the quadratic formula, represented by \(b^2 - 4ac\). It is crucial because it tells you about the number and type of solutions a quadratic equation has.

Here are the key points:

• If the discriminant is positive (greater than zero), you will have two distinct real solutions. This means the parabola represented by the equation crosses the x-axis at two points.
• If the discriminant is zero, you will have exactly one real solution. This implies that the parabola just touches the x-axis and is known as a repeated or double root.
• If the discriminant is negative, the equation has no real solutions but two complex solutions, as the parabola does not intersect the x-axis.

In our original problem, the discriminant is calculated as 16, indicating two distinct real solutions are possible. This sets the stage for finding the precise values of \(x\) when plugged back into the quadratic formula.

Algebraic Solutions

Using algebraic methods to solve quadratic equations involves finding exact values of \(x\) that satisfy the equation. Once you've determined the discriminant is positive, you can use the quadratic formula to find the solutions. As seen in our step-by-step process:

1. Calculate the discriminant: \(b^2 - 4ac\). This needs to be computed so that you know whether real solutions exist.
2. Plug values into the quadratic formula: After finding the discriminant, use \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to find \(x_1\) and \(x_2\).
3. Simplify your expressions: Simplifying involves performing square root operations and basic arithmetic like addition, subtraction, multiplication, and division.

In our example, following the substitution, we found two solutions: \(x_1 = \frac{5}{6}\) and \(x_2 = \frac{1}{2}\), meaning these are the x-values where the parabola meets the x-axis. Checking these in the original equation ensures they are correct. Understanding each step thoroughly helps to achieve an accurate solution and reinforces the skill of algebraic problem-solving.