Problem 53 - Intermediate Algebra

Question

Solve each quadratic equation using the method that seems most appropriate. $$ 3 x^{2}+5 x=-2 $$

Step-by-Step Solution

Answer

The solutions are \(-\frac{2}{3}\) and \(-1\).

Step 1: Bring Equation into Standard Form

To use any method for solving a quadratic equation, we need it in standard form: \( ax^2 + bx + c = 0 \). The given equation is \( 3x^2 + 5x = -2 \). Add 2 to both sides to get: \( 3x^2 + 5x + 2 = 0 \).

Step 2: Identify Coefficients

Identify the coefficients from the standard form \( ax^2 + bx + c = 0 \). Here, \( a = 3 \), \( b = 5 \), and \( c = 2 \).

Step 3: Use the Quadratic Formula

The quadratic formula is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute in the coefficients: \( x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} \).

Step 4: Simplify Inside the Square Root

Calculate \( b^2 - 4ac \): \( 5^2 - 4 \cdot 3 \cdot 2 = 25 - 24 = 1 \). Now, substitute back: \( x = \frac{-5 \pm \sqrt{1}}{6} \).

Step 5: Compute the Roots

Since \( \sqrt{1} = 1 \), the equation becomes \( x = \frac{-5 \pm 1}{6} \). This gives two solutions: \( x = \frac{-5 + 1}{6} = \frac{-4}{6} = -\frac{2}{3} \) and \( x = \frac{-5 - 1}{6} = \frac{-6}{6} = -1 \).

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool for solving quadratic equations. These equations take the form \( ax^2 + bx + c = 0 \). The formula is used to find the values of \( x \) that satisfy the equation, commonly known as the "roots" of the equation. The quadratic formula is expressed as:

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

This formula allows you to calculate the roots of any quadratic equation by plugging in values for \( a \), \( b \), and \( c \) from the equation. It's especially useful when the quadratic equation is not easily factorable.
Solving quadratic equations using this formula involves:

• Substituting the coefficients \( a \), \( b \), and \( c \) into the formula.
• Calculating the "discriminant," which is the expression under the square root: \( b^2 - 4ac \).
• Determining the square root of the discriminant.
• Applying the formula's plus-minus (\( \pm \)) symbol to find two potential root values.

For example, in the equation \( 3x^2 + 5x + 2 = 0 \), the coefficients are \( a = 3 \), \( b = 5 \), \( c = 2 \). Plugging these into the quadratic formula helps us find the roots of the equation.

Standard Form of Quadratic Equation

The standard form of a quadratic equation provides a consistent and clear way to express quadratic equations. This form is written as \( ax^2 + bx + c = 0 \), where:

• \( a \), \( b \), and \( c \) are constants.
• \( a \) is not zero (otherwise the equation becomes linear, not quadratic).
• \( x \) represents the variable to be solved for.

Getting a quadratic equation into this form is important for solving it using various methods, like factoring, completing the square, or using the quadratic formula. Each method requires the equation to be arranged with zero on one side.
In our original exercise, the equation started as \( 3x^2 + 5x = -2 \). By adding 2 to both sides, we convert it into standard form: \( 3x^2 + 5x + 2 = 0 \). This standard format made it possible to identify coefficients \( a \), \( b \), and \( c \) easily, which is crucial for applying the quadratic formula.

Roots of Quadratic Equation

Roots of a quadratic equation are the solutions to the equation \( ax^2 + bx + c = 0 \). These roots are the values of \( x \) that make the equation true. A quadratic equation typically has two roots, which can be real or complex numbers, depending on the discriminant.
The discriminant, \( b^2 - 4ac \), determines the nature of the roots.

• If the discriminant is positive, the equation has two distinct real roots.
• If it is zero, the equation has exactly one real root (a repeated root).
• If negative, the roots are complex and occur as a conjugate pair.

In our example, \( b^2 - 4ac = 1 \) is a positive number. This tells us the equation has two distinct real roots. Solving \( 3x^2 + 5x + 2 = 0 \) using the quadratic formula, we found the roots: \( x = -\frac{2}{3} \) and \( x = -1 \), meaning at these values of \( x \), the original equation equals zero.