Problem 38 - Algebra A Combined Function

Question

Use the quadratic formula to solve each equation. These equations have real solutions and complex, but not real, solutions. $$ \frac{2}{3} x^{2}-\frac{20}{3} x=-\frac{100}{6} $$

Step-by-Step Solution

Answer

The solution is \( x = 5 \).

Step 1: Rearrange the equation

First, rearrange the equation to standard quadratic form \( ax^2 + bx + c = 0 \). The given equation is \( \frac{2}{3}x^{2} - \frac{20}{3}x = -\frac{100}{6} \). Multiply every term by 6 to clear the denominators, resulting in \( 4x^2 - 40x + 100 = 0 \).

Step 2: Identify coefficients

Identify the coefficients from the standard form \( ax^2 + bx + c = 0 \). In \( 4x^2 - 40x + 100 = 0 \), the coefficients are: \( a = 4 \), \( b = -40 \), and \( c = 100 \).

Step 3: Write the quadratic formula

Recall the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula is used to find the solutions of a quadratic equation.

Step 4: Calculate the discriminant

Calculate the discriminant \( \Delta \) using \( \Delta = b^2 - 4ac \). Substitute the values: \( \Delta = (-40)^2 - 4 \times 4 \times 100 \). Simplifying gives \( \Delta = 1600 - 1600 = 0 \).

Step 5: Evaluate solutions with the quadratic formula

Since the discriminant is zero, there is exactly one real solution. Substitute into the quadratic formula: \( x = \frac{-(-40) \pm \sqrt{0}}{2 \times 4} \). This simplifies to \( x = \frac{40}{8} = 5 \). The only solution is \( x = 5 \).

Key Concepts

Discriminant

The discriminant is a crucial component of the quadratic formula, often symbolized by \( \Delta \). It is used to determine the nature and number of roots for a quadratic equation. To calculate the discriminant, use the formula:

• \( \Delta = b^2 - 4ac \)

In this expression, \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation. Determining the value of the discriminant reveals the type of solutions:

• If \( \Delta > 0 \), there are two distinct real solutions.
• If \( \Delta = 0 \), there is exactly one real solution (also known as a repeated root).
• If \( \Delta < 0 \), there are no real solutions, but two complex solutions.

In our provided equation example, \( \Delta = 0 \), indicating a single real solution.

Quadratic Equation

A quadratic equation is any equation that can be structured in the form \( ax^2 + bx + c = 0 \). This configuration is known as the standard form of a quadratic equation, where:

• \( a \), \( b \), and \( c \) are constants.
• \( x \) is the variable.
• \( a \) must not be equal to zero.

Quadratic equations often result in one or two solutions and are foundational in algebra due to their diverse applications in fields such as physics, engineering, and economics.To solve a quadratic equation, one widely used method is the quadratic formula:

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

To apply this formula successfully, the equation must first be converted into its standard form. This process involves rearranging and simplifying the equation so that it aligns with \( ax^2 + bx + c = 0 \). In the example exercise, the initial equation was transformed to \( 4x^2 - 40x + 100 = 0 \).

Real Solutions

The term "real solutions" refers to solutions of a quadratic equation that are real numbers. Unlike complex solutions, real solutions do not involve imaginary numbers. They are solutions that lie on the real number line.The nature of real solutions in the context of a quadratic equation is largely determined by the discriminant:

• If the discriminant \( \Delta > 0 \), the equation has two unique real solutions.
• If \( \Delta = 0 \), the equation has one real solution; this is known as a repeated or double root.
• If \( \Delta < 0 \), the equation does not have real solutions but instead has complex solutions.

For real-world scenarios, such as calculating projectile motion or finding optimal values in business, real solutions are extensively utilized as they provide concrete values that can be applied to practical problems. In our problem, with a discriminant of zero, the equation yields exactly one real solution, which we determined to be \( x = 5 \). This solution represents the point where the parabola defined by the quadratic equation touches the x-axis.