Problem 31 - Algebra for College Students

Question

Use the quadratic formula to solve each of the quadratic equations. Check your solutions by using the sum and product relationships. $$3 x^{2}+19 x+20=0$$

Step-by-Step Solution

Answer

The solutions are \( x = -\frac{4}{3} \) and \( x = -5 \).

Step 1: Identify coefficients

The given equation is in the form \( ax^2 + bx + c = 0 \). Here, \( a = 3 \), \( b = 19 \), and \( c = 20 \). Identify these coefficients as the first step before applying the quadratic formula.

Step 2: Apply the quadratic formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute the values of \( a \), \( b \), and \( c \) into this formula to find the solutions. Here, we have \( x = \frac{-19 \pm \sqrt{19^2 - 4 \cdot 3 \cdot 20}}{2 \cdot 3} \).

Step 3: Calculate the discriminant

Calculate the discriminant, \( b^2 - 4ac \). In this problem, it becomes \( 19^2 - 4 \cdot 3 \cdot 20 = 361 - 240 = 121 \).

Step 4: Determine the square root of the discriminant

The square root of the discriminant, \( \sqrt{121} \), is \( 11 \). Use this result in the quadratic formula.

Step 5: Calculate the solutions

Substitute \( \sqrt{121} = 11 \) back into the quadratic formula to get the solutions: \( x = \frac{-19 \pm 11}{6} \). This results in two solutions: \( x = \frac{-19 + 11}{6} = \frac{-8}{6} = -\frac{4}{3} \) and \( x = \frac{-19 - 11}{6} = \frac{-30}{6} = -5 \).

Step 6: Verify solutions using sum and product relationships

For a quadratic equation \( ax^2 + bx + c = 0 \), the sum of roots \( \alpha + \beta \) is \( -\frac{b}{a} \), and the product of roots \( \alpha \cdot \beta \) is \( \frac{c}{a} \). Here, the sum is \( -\frac{19}{3} \) and the product is \( \frac{20}{3} \). Verify with solutions: \(-\frac{4}{3} + (-5) = -\frac{4}{3} - \frac{15}{3} = -\frac{19}{3}\). The product: \( -\frac{4}{3} \cdot (-5) = \frac{20}{3} \). Both checks are correct.

Key Concepts

Quadratic Formula

The quadratic formula is a universal method for solving quadratic equations of the form \( ax^2 + bx + c = 0 \). It allows you to find the roots, or solutions, of the equation. The formula is expressed as:

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

To use the quadratic formula, first identify the coefficients \( a \), \( b \), and \( c \) from your equation. In our given example \( 3x^2 + 19x + 20 = 0 \), we have:

• \( a = 3 \)
• \( b = 19 \)
• \( c = 20 \)

After plugging these values into the quadratic formula, you calculate the part under the square root, called the discriminant (more on this later), and solve for \( x \). Remember that the formula provides two possible solutions, using \( + \) or \( - \) in front of the square root.

Discriminant

The discriminant is a key part of the quadratic formula, found within the square root: \( b^2 - 4ac \). It determines the nature of the roots without actually solving the entire formula. Here's why it's important:

• If the discriminant is positive, you get two distinct real roots.
• If the discriminant is zero, there is exactly one real root, known as a repeated root.
• If the discriminant is negative, there are no real roots; instead, the solutions are complex numbers.

In our exercise, the discriminant calculation is:

• \( 19^2 - 4 \times 3 \times 20 = 361 - 240 = 121 \)

Since 121 is positive, this indicates two distinct real roots. The discriminant isn't only a number; it's the key to understanding how many solutions exist. In practical terms, you always solve for the discriminant before proceeding to find the square root value and completing the quadratic formula.

Sum and Product of Roots

Beyond just finding the roots using the quadratic formula, the relationships between the quadratic equation's coefficients and its roots provide valuable insight. These relationships are known as the sum and product of roots:

• The sum of the roots, \( \alpha + \beta \), is given by \( -\frac{b}{a} \).
• The product of the roots, \( \alpha \cdot \beta \), is \( \frac{c}{a} \).

For the equation \( 3x^2 + 19x + 20 = 0 \), the sum and product can be directly calculated:

• Sum: \( -\frac{19}{3} \)
• Product: \( \frac{20}{3} \)

After solving with the quadratic formula, we find the roots \( -\frac{4}{3} \) and \( -5 \). Check these using:

• Sum: \( -\frac{4}{3} + (-5) = -\frac{19}{3} \)
• Product: \( -\frac{4}{3} \times (-5) = \frac{20}{3} \)

These values match the calculations, confirming the correctness of our solutions. Understanding these relationships not only helps verify your answers, but also deepens your grasp of quadratic equations.