A quadratic equation is a polynomial equation of the second degree. It is in the form of ax^2 + bx + c = 0, where a, b, and c are real numbers and a ≠ 0.
Quadratic equations can be solved using a variety of methods, including factoring, completing the square, and the quadratic formula. This guide will cover all three methods in detail, and provide examples of how to use each method to solve quadratic equations.
Method 1: Factoring
Factoring is the process of breaking down a polynomial into smaller, more manageable polynomials. Quadratic equations can be factored into two linear equations, ax + b = 0 and cx + d = 0. Once the quadratic equation has been factored, the two linear equations can be solved independently to find the roots of the quadratic equation.
To factor a quadratic equation, you can use a variety of methods, such as trial and error, grouping, and perfect square trinomials.
Example:
Factor the following quadratic equation:
x^2 + 5x + 6 = 0
This quadratic equation can be factored into two linear equations:
(x + 3)(x + 2) = 0
To solve for the roots of the quadratic equation, we set each linear equation equal to zero and solve for x.
x + 3 = 0
x = -3
x + 2 = 0
x = -2
Therefore, the roots of the quadratic equation x^2 + 5x + 6 = 0 are x = -3 and x = -2.
Method 2: Completing the Square
Completing the square is a method of solving quadratic equations by adding a constant term to both sides of the equation so that the left-hand side of the equation becomes a perfect square trinomial. Once the left-hand side of the equation is a perfect square trinomial, it can be factored and solved using the square root property.
To complete the square, you first need to divide both sides of the equation by the leading coefficient (the coefficient of the x^2 term). Then, you need to add half of the coefficient of the x term, squared, to both sides of the equation.
Example:
Solve the following quadratic equation using completing the square:
x^2 - 4x - 5 = 0
First, divide both sides of the equation by 1, the leading coefficient:
x^2 - 4x = 5
Next, add half of the coefficient of the x term, squared, to both sides of the equation. The coefficient of the x term is -4, so half of it would be -2, and the square of -2 is 4.
x^2 - 4x + 4 = 5 + 4
(x - 2)^2 = 9
Now, factor the left-hand side of the equation and solve using the square root property:
√(x - 2)^2 = √9
x - 2 = ±3
x = 2 ± 3
Therefore, the roots of the quadratic equation x^2 – 4x – 5 = 0 are x = 5 and x = -1.
Method 3: The Quadratic Formula
The quadratic formula is a general formula that can be used to solve any quadratic equation. It is derived from the factoring and completing the square methods.
The quadratic formula is given by the following equation:
x = (-b ± √(b^2 - 4ac)) / 2a
where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0.
Example:
Solve the following quadratic equation using the quadratic formula:
x^2 + 3x - 4 = 0
In this equation, a = 1, b = 3, and c = -4. Substituting these values into the quadratic formula, we get the following:
x = (-3 ± √(3^2 - 4 * 1 * -4)) / 2 * 1
x = (-3 ± √25) / 2
x = (-3 ± 5) / 2
Therefore, the roots of the quadratic equation x^2 + 3x – 4 = 0 are x = 1 and x = -4.
Applications of Quadratic Equations
Quadratic equations have many real-world applications. For example, they can be used to model the motion of a projectile, the growth of a population, and the profit of a business.
Example:
A projectile is launched into the air with an initial velocity of 100 m/s. The height of the projectile at any given time t is given by the following quadratic equation:
h(t) = -4.9t^2 + 100t
To find the time it takes for the projectile to reach its maximum height, we need to solve the following quadratic equation:
h(t) = 0
-4.9t^2 + 100t = 0
t(4.9 - 20) = 0
t = 0 or t = 4.08 seconds
Therefore, the projectile reaches its maximum height after 4.08 seconds.
Conclusion
Quadratic equations can be solved using a variety of methods, including factoring, completing the square, and the quadratic formula. Each method has its own advantages and disadvantages. The best method to use depends on the specific quadratic equation.
