Problem 57 - Elementary Algebra

Question

Use the quadratic formula to solve the following. $$3 y 2+12 y-13=0$$

Step-by-Step Solution

Answer

The solutions are \( x_1 = -2 + \frac{5\sqrt{3}}{3} \) and \( x_2 = -2 - \frac{5\sqrt{3}}{3} \).

Step 1: Identify the coefficients

To use the quadratic formula, first identify the coefficients in the standard quadratic form, which is \( ax^2 + bx + c = 0 \). For the equation \(3y^2 + 12y - 13 = 0\), the coefficients are: \( a = 3 \), \( b = 12 \), and \( c = -13 \).

Step 2: Write the quadratic formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula will help us calculate the solutions for the quadratic equation.

Step 3: Substitute the coefficients into the formula

Replace \( a \), \( b \), and \( c \) in the quadratic formula with the identified values: \( x = \frac{-12 \pm \sqrt{12^2 - 4 \times 3 \times (-13)}}{2 \times 3} \).

Step 4: Calculate the discriminant

Calculate the discriminant \( b^2 - 4ac \): \( 12^2 - 4 \times 3 \times (-13) = 144 + 156 = 300 \).

Step 5: Solve under the square root

Simplify the square root part of the equation: \( \sqrt{300} \). This can be simplified further noting that \( 300 = 100 \times 3 \), so \( \sqrt{300} = 10\sqrt{3} \).

Step 6: Simplify the expression

Now substitute back: \( x = \frac{-12 \pm 10\sqrt{3}}{6} \). Divide each term in the numerator by 6: \( x = \frac{-12}{6} \pm \frac{10\sqrt{3}}{6} = -2 \pm \frac{5\sqrt{3}}{3} \).

Step 7: State the final solutions

The solutions are \( x_1 = -2 + \frac{5\sqrt{3}}{3} \) and \( x_2 = -2 - \frac{5\sqrt{3}}{3} \).

Key Concepts

Quadratic Equations

Quadratic equations are an important class of polynomial equations that have a degree of two. This means they include terms with the variable raised to powers up to two. The general form of a quadratic equation is \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). They are widely used in various scientific and engineering problems because they can model numerous situations.

The solutions to quadratic equations are the values of the variable that satisfy the equation. These solutions can be real or complex numbers, depending on various factors which we will discuss further. Solving quadratic equations typically involves factoring, completing the square, or using the quadratic formula, which is particularly handy due to its straightforward application regardless of the specific numbers involved. Understanding the structure of quadratic equations is essential for solving them effectively.

Discriminant

The discriminant is a component of the quadratic formula that gives insight into the nature of the roots of the quadratic equation. It is denoted as \( b^2 - 4ac \) from the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). The value of the discriminant tells us if the roots are real or complex, and if they are distinct or repeated.

• If the discriminant is positive, the quadratic equation has two distinct real roots. For example, a discriminant of 300 indicates two distinct real roots.
• If the discriminant is zero, there is exactly one real root, repeated. This means the parabola touches the x-axis at one point.
• If the discriminant is negative, the solutions are complex conjugates, which means the graph of the quadratic will not intersect the x-axis.

Understanding the discriminant is key to predicting the nature of the solutions without fully solving the equation. In our exercise, the discriminant was 300, pointing to two distinct real roots.

Solving Quadratics

To solve quadratic equations, one of the most versatile tools is the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This formula provides a method to find solutions directly from the coefficients of the quadratic equation. Let’s break down the solving process:

• First, identify the coefficients \( a \), \( b \), and \( c \) from the given equation.
• Use these values to find the discriminant \( b^2 - 4ac \).
• Calculate \( \sqrt{b^2 - 4ac} \) and substitute it into the formula.

In our example, by substituting \( a = 3 \), \( b = 12 \), and \( c = -13 \) into the formula, we calculated the discriminant as 300. We then continued to simplify under the square root, yielding \( 10\sqrt{3} \). Finally, by dividing the entire expression by \( 2a \), we obtained the two solutions: \(-2 + \frac{5\sqrt{3}}{3}\) and \(-2 - \frac{5\sqrt{3}}{3} \).

This systematic process allows us to find the roots of any quadratic equation, no matter how complicated it looks initially. The quadratic formula is a reliable method because it simplifies and accommodates equations that are not easily factored or completed by square.