Solving Quadratic Equations Using the Quadratic Formula

Introduction to the Quadratic Formula

The primary objective when solving a quadratic equation is to calculate the specific value or values of the variable $x$ that satisfy the equation, making the mathematical statement true. A fundamental tool for achieving this is the quadratic formula.

Standard Form and Coefficient Identification

Before applying the quadratic formula, the quadratic equation must be arranged in standard form. This requires all variables and constant terms to be positioned on the left side of the equality, with zero representing the right side. In this configuration, the coefficients are identified as follows:

• $a$ corresponds to the coefficient in front of the term $x^2$.

• $b$ corresponds to the coefficient in front of the term $x$.

• $c$ corresponds to the constant term (the number without a variable).

The Quadratic Formula Expression

The formula used to determine the roots of a quadratic equation is expressed as:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

This formula accommodates the possibility of two distinct solutions through the use of the plus or minus ($\pm$) symbol.

Procedural Breakdown: Case Study 1

Consider the quadratic equation:

$2x^2 + 3x - 2 = 0$

Step 1: Identify Coefficients

From the equation, the following values are extracted:

• $a = 2$

• $b = 3$

• $c = -2$

Step 2: Substitution into the Formula

Substitute the identified values into the quadratic formula:

$x = \frac{-(3) \pm \sqrt{(3)^2 - 4(2)(-2)}}{2(2)}$

Step 3: Simplify the Expression

Evaluate the components within the formula:

• The numerator's leading term is $-3$.

• The term $b^2$ is $3 \times 3 = 9$.

• The product of $-4ac$ is calculated as $-4 \times 2 = -8$, and $-8 \times -2 = 16$.

• The denominator is $2 \times 2 = 4$.

Combining the radicand values:

$9 + 16 = 25$

This results in the simplified expression:

$x = \frac{-3 \pm \sqrt{25}}{4}$

Step 4: Evaluate the Square Root

Since the square root of $25$ is $5$, the expression becomes:

$x = \frac{-3 \pm 5}{4}$

Step 5: Solve for Both Possible Values

Separate the expression into two distinct calculations based on the plus or minus symbol:

1. For the plus case:     $x = \frac{-3 + 5}{4} = \frac{2}{4} = \frac{1}{2}$

2. For the minus case:     $x = \frac{-3 - 5}{4} = \frac{-8}{4} = -2$

Verification of Solutions

To ensure the accuracy of the calculated roots, the values can be substituted back into the original equation.

Testing $x = -2$ in the equation $2x^2 + 3x - 2 = 0$:

$2(-2)^2 + 3(-2) - 2 = 0$

$2(4) - 6 - 2 = 0$

$8 - 6 - 2 = 0$

$0 = 0$

Since the left side equals the right side, the answer $x = -2$ is verified. Substituting $1/2$ would yield a similar result.

Procedural Breakdown: Case Study 2

Consider a second equation with different coefficients:

$6x^2 - 17x + 12 = 0$

Step 1: Identify Coefficients

• $a = 6$

• $b = -17$

• $c = 12$

Step 2: Substitution

When $b$ is negative, the term $-b$ in the formula becomes positive:

$x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4(6)(12)}}{2(6)}$

Step 3: Arithmetic Simplification

• $-(-17)$ becomes $17$.

• $(-17)^2 = 289$.

• The product $-4 \times 6 \times 12$ is calculated as $-24 \times 12 = -288$.

• The denominator is $2 \times 6 = 12$.

$x = \frac{17 \pm \sqrt{289 - 288}}{12}$

$x = \frac{17 \pm \sqrt{1}}{12}$

Step 4: Evaluate the Square Root

Since the square root of $1$ is $1$, the equation simplifies further:

$x = \frac{17 \pm 1}{12}$

Step 5: Solve for Both Fractions

Divide the problem into two distinct parts:

1. First solution:     $x = \frac{17 + 1}{12} = \frac{18}{12}$     Reduce the fraction by identifying common factors ($6 \times 3 = 18$ and $6 \times 2 = 12$):     $x = \frac{3}{2}$

2. Second solution:     $x = \frac{17 - 1}{12} = \frac{16}{12}$     Reduce the fraction by identifying common factors ($4 \times 4 = 16$ and $4 \times 3 = 12$):     $x = \frac{4}{3}$

The final solutions for the equation are $3/2$ and $4/3$.