2-2-3 Quadratic Equations Explained

Key Concepts of Quadratic Equations

Quadratic equations are polynomial equations of the second degree. They are expressed in the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \neq 0 \). The solutions to a quadratic equation are called the roots or zeros of the equation.

1. Standard Form of a Quadratic Equation

The standard form of a quadratic equation is:

\[ ax^2 + bx + c = 0 \]

Here, \( a \) is the coefficient of \( x^2 \), \( b \) is the coefficient of \( x \), and \( c \) is the constant term.

Example:

The equation \( 2x^2 + 3x - 5 = 0 \) is in standard form, where \( a = 2 \), \( b = 3 \), and \( c = -5 \).

2. Solving Quadratic Equations

There are several methods to solve quadratic equations:

Factoring: If the quadratic equation can be factored, it is often the simplest method.
Completing the Square: This method involves rewriting the quadratic equation in a form that allows for easy solving.
Quadratic Formula: This formula provides the solutions directly and is applicable to all quadratic equations.

Factoring

To factor a quadratic equation, you need to find two numbers that multiply to \( ac \) and add to \( b \). Then, rewrite the equation as a product of two binomials.

Example:

Solve \( x^2 + 5x + 6 = 0 \):

Factor the equation: \( (x + 2)(x + 3) = 0 \)

Set each factor to zero: \( x + 2 = 0 \) or \( x + 3 = 0 \)

Solve for \( x \): \( x = -2 \) or \( x = -3 \)

Completing the Square

This method involves transforming the quadratic equation into a perfect square trinomial.

Example:

Solve \( x^2 + 6x + 5 = 0 \):

Move the constant term to the other side: \( x^2 + 6x = -5 \)

Add and subtract \((\frac{b}{2})^2\): \( x^2 + 6x + 9 = -5 + 9 \)

Rewrite as a perfect square: \( (x + 3)^2 = 4 \)

Solve for \( x \): \( x + 3 = \pm 2 \)

So, \( x = -1 \) or \( x = -5 \)

Quadratic Formula

The quadratic formula is:

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

This formula can be used to find the roots of any quadratic equation.

Example:

Solve \( 2x^2 + 3x - 5 = 0 \):

Identify \( a = 2 \), \( b = 3 \), and \( c = -5 \).

Substitute into the quadratic formula:

\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-5)}}{2 \cdot 2} \]

\[ x = \frac{-3 \pm \sqrt{9 + 40}}{4} \]

\[ x = \frac{-3 \pm \sqrt{49}}{4} \]

\[ x = \frac{-3 \pm 7}{4} \]

So, \( x = 1 \) or \( x = -\frac{5}{2} \)

3. Discriminant and Nature of Roots

The discriminant of a quadratic equation \( ax^2 + bx + c = 0 \) is given by \( \Delta = b^2 - 4ac \). The value of the discriminant determines the nature of the roots:

If \( \Delta > 0 \): The equation has two distinct real roots.
If \( \Delta = 0 \): The equation has exactly one real root (a repeated root).
If \( \Delta < 0 \): The equation has two complex (non-real) roots.

Example:

For the equation \( x^2 + 4x + 4 = 0 \):

Calculate the discriminant: \( \Delta = 4^2 - 4 \cdot 1 \cdot 4 = 16 - 16 = 0 \)

Since \( \Delta = 0 \), the equation has exactly one real root.

4. Applications of Quadratic Equations

Quadratic equations are used in various real-world applications, such as:

Calculating the trajectory of projectiles.
Determining the maximum or minimum value of a function.
Solving problems in physics and engineering.

Example:

A ball is thrown upwards with an initial velocity of 20 m/s. The height \( h \) of the ball after \( t \) seconds is given by the equation \( h = -4.9t^2 + 20t \). Find the time when the ball hits the ground.

Set \( h = 0 \): \( -4.9t^2 + 20t = 0 \)

Factor the equation: \( t(-4.9t + 20) = 0 \)

Solve for \( t \): \( t = 0 \) or \( t = \frac{20}{4.9} \approx 4.08 \) seconds.