Problem 29 - Algebra A Combined Function

Question

Use the quadratic formula to solve each equation. These equations have real solutions and complex, but not real, solutions. $$ \frac{x^{2}}{3}-x=\frac{5}{3} $$

Step-by-Step Solution

Answer

The real solutions are \( x = \frac{3 + \sqrt{29}}{2} \) and \( x = \frac{3 - \sqrt{29}}{2} \).

Step 1: Clear Fractions by Multiplying through by the Denominator

Start with the given equation: \( \frac{x^2}{3} - x = \frac{5}{3} \). Multiply each term by 3 to eliminate the fractions, resulting in \( x^2 - 3x = 5 \).

Step 2: Rearrange Equation into Standard Form

Rearrange the equation into the standard quadratic form \( ax^2 + bx + c = 0 \). This gives us \( x^2 - 3x - 5 = 0 \). Here, \( a = 1 \), \( b = -3 \), and \( c = -5 \).

Step 3: Recall the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). We will use this formula to find the solutions of the quadratic equation.

Step 4: Calculate the Discriminant

Calculate the discriminant \( b^2 - 4ac \) using the values \( a = 1 \), \( b = -3 \), and \( c = -5 \). Substitute these into the formula: \( (-3)^2 - 4(1)(-5) = 9 + 20 = 29 \). Since the discriminant is positive, the solutions will be real and distinct.

Step 5: Use the Quadratic Formula to Find Solutions

Substitute \( a = 1 \), \( b = -3 \), and the calculated discriminant into the quadratic formula: \( x = \frac{-(-3) \pm \sqrt{29}}{2(1)} = \frac{3 \pm \sqrt{29}}{2} \).

Step 6: Interpret the Solutions

The solutions to the quadratic equation are \( x = \frac{3 + \sqrt{29}}{2} \) and \( x = \frac{3 - \sqrt{29}}{2} \). Both solutions are real and distinct.

Key Concepts

Discriminant

In the world of quadratic equations, the discriminant is a key concept that helps determine the nature of the solutions. The discriminant is denoted as \( b^2 - 4ac \), where \( a \), \( b \), and \( c \) are the coefficients of the equation in its standard form, \( ax^2 + bx + c = 0 \).
Here's how to interpret the discriminant:

• If the discriminant is positive, the quadratic equation has two distinct real solutions.
• If the discriminant is zero, the equation has exactly one real solution, which is repeated.
• If the discriminant is negative, the solutions are complex and not real.

In our exercise, the discriminant was calculated as 29, which is positive. Therefore, the equation has two distinct real solutions. This information gives us a heads-up about what to expect before solving the equation fully using the quadratic formula.

Quadratic Equations

Quadratic equations are fundamental components of algebra. A quadratic equation is any equation that can be rearranged in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).
This is the standard form, and our goal is often to find the values of \( x \) that satisfy the equation. These values are known as the roots or solutions.
To solve such equations, various methods can be used, but the quadratic formula is a universal tool: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] The plus-minus sign (\( \pm \)) in the formula indicates that there are usually two solutions. These solutions could be real or complex, depending on the value of the discriminant. In our case, after clearing fractions and rearranging the equation to \( x^2 - 3x - 5 = 0 \), we identified \( a = 1 \), \( b = -3 \), and \( c = -5 \) to apply the quadratic formula.

Clear Fractions

Clearing fractions in an equation is often a crucial first step. It simplifies calculations and avoids errors in computation.
For equations involving fractions, it's helpful to eliminate them by multiplying each term by the least common denominator (LCD).
For instance, in the original equation \( \frac{x^2}{3} - x = \frac{5}{3} \), we multiplied every term by 3. This gives us \( x^2 - 3x = 5 \), making it easier to rearrange into the quadratic form.
The main purpose of clearing fractions is to prevent dealing with complicated arithmetic, which can obscure the simple structure of the quadratic equation once fractions are removed. Once fractions are cleared, it becomes straightforward to apply further algebraic techniques such as completing the square or using the quadratic formula.