A quadratic equations is a polynomial equations of degree 2. The 'U' shaped graph of a quadratic is called a parabola. A quadratic equation has two solutions. Either two distinct real solutions, one double real solution or two imaginary solutions. Solving Quadratic Equations

There are four main ways to solve quadratic equations:

1. to factor the quadratic equation,
2. to use the quadratic formula,
3. to complete the square,
4. graphing.

If you want to know how to master these four methods, just follow these steps.

## Solving Quadratic Equations By Factoring

Example 1:
Solve for x in the following equation: x2 + 6x + 8 = 0.
This equation is already in the form "(quadratic) equals (zero)" but, this isn't yet factored. The quadratic must first be factored, because it is only when you MULTIPLY and get zero that you can say anything about the factors and solutions. You can't conclude anything about the individual terms of the unfactored quadratic (like the 5x or the 6), because you can add lots of stuff that totals zero.
So the first thing I have to do is factor:

x2 + 6x + 8 = (x + 2)(x + 4) Set this equal to zero:
(x + 2)(x + 4) = 0 Solve each factor:
x + 2 = 0 or x + 4 = 0
x = –2 or x = – 4
The solution to x2 + 6x + 8 = 0 is x = (–4, –2).

Checking x = –4 and x = –2 in x2 + 6x + 8 = 0:
[–4]2 + 6[–4] + 8 = 0
16 – 24 + 8 = 0
16 + 8 – 24 = 0
24 – 24 = 0
0 = 0.

[–2]2 + 6[–2] + 8 = 0
4 – 12 + 8 = 0
4 + 8 – 12 = 0
12 – 12 = 0
0 = 0.

So both solutions "check".

Example 2:
Solve x2 – 3 = 2x. This equation is not in "quadratic equals zero" form, so you can't try to solve it yet. The first thing you need to do is get all the terms over on one side, with zero on the other side. Only then can you factor and solve:

x2 – 3 = 2x
x2 – 2x – 3 = 0
(x – 3)(x + 1) = 0
x – 3 = 0 or x + 1 = 0
x = 3 or x = –1
Then the solution to x2 – 3 = 2x is x = (–1, 3).
Checking...etc.

## Solving Quadratic Equation by Completing the Square

The method of 'completing the square' can be applied to quadratic equations that are written in the standard form. This method uses the algebraic property of expansion of the square of the sum of two terms, which means:
(a + b)2 = a2 +2ab + b2

In this method, simple mathematical operations are applied to a quadratic equation (written in the standard form) in order to make its LHS (Left Hand Side) expression of the form a2  + 2ab + b2 . Then the LHS expression is factored to the form (a + b)2  by applying the above mentioned property.

The following example will further clarify this method:

Example 1:
Solve x2  + 4x - 5 = 0 !
ax2  +bx + c = 0

1) Move the constant to the right side. In the equation above, the constant is 5. You can do this by adding the negative value of the constant on both sides of the equation;
x2  + 4x -5 + (+5) = 0 + (+5)
x2  + 4x = 5

2) Take half the value of the x-term or b (4).
4/2 = 2

3) Square it.
22  = 4

4) Add the result to both sides.
x2  + 4x + 4 = 5 + 4

5) Convert the left side to squared form and simplify the right side.
( x + 2)2  = 9

6) Take the square root of both sides.
x + 2 = +/- 3

7) Solve for x
x + 2 = 3
x = 3 - 2
x = 1
or,
x + 2 = -3
x = -2 - 3
x = -5
x = 1 and x = -5.

For ax2 + bx + c = 0, the value of x is given by:

This formula will work for all equations that can be solved.
Always try to factorise first. If the equation factorises, this is the easier method. In an exam, any question that asks for an answer to a quadratic equation correct to x decimal places should be solved using this formula. Take a look at this example:
Example 1:
Solve 2x2 - 5x - 6 = 0
Here a = 2, b = -5, c = -6

Substituting these values in the formula, gives: Related: Solve Systems of Linear Equations in Two Variables

## Solving Quadratic Equations by Graphing

Solving quadratic equations by graphing step by step:Graph y= the left side of the equation or y = x2 -x-1 and graph y= the right side of the equation or y=0. The graph of y=0 is nothing more than the x-axis. So what you will be looking for is where the graph of y = 2x2 -x -1 crosses the x-axis. Another way of saying this is that the x-intercepts are the solutions to this equation.
You can see from the graph that there are two x-intercepts, one at 1 and one at -1/2 .

The answers are 1 and -1/2, These answers may or may not be solutions to the original equations. You must verify that these answers are solutions.

Check these answers in the original equation.
Check the solution x=1 by substituting 1 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
Left Side:2(1)2 - (1) - 1 = 0
Right Side: 0

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value 1 for x, then x=1 is a solution.

Check the solution  x = -1/2 by substituting -1/2 in the original equation for x. If the left side of the equation equals the right side of the equation after the substitution, you have found the correct answer.
Left Side: 2 (-1/2)2 - (-1/2) - 1 = 0
Right Side: 0

Since the left side of the original equation is equal to the right side of the original equation after we substitute the value -1/2 for x, then x = -1/2 is a solution.

The solutions to the equation 2x2 -x -1 = 0 are 1 and -1/2.