Three Ways Of Solving Quadratic Equations

Three Ways to Solve Quadratic Equations
How do I solve this?

x^2 + 4x + 2 = 0



What you have there is a quadratic equation: an equation with a single
variable where the largest poweris a 2.

There are three main ways to solve quadratic equations:

* by inspection - Note: this does not work for every quadratic
* by completing the square
* by using the quadraticformula

Which method you end up using is up to you; everyone has his or her
own preference. I suggest trying each of the methods with your
question and seeing which you prefer. I will use x^2 - x =20 as an
example for all three.

Before starting any of the methods, you need to use addition and
subtraction to get everything on one side of the equals sign.

x^2 - x = 20
x^2 -x - 20 = 0


SOLVING BY INSPECTION:
(Some people know this as solving by factoring.)

x^2 - x - 20 = 0

First you factorise the equation into a format like:

(ax + c)(bx + d) = 0This looks a bit complicated, but it isn't really. Multiply out the
parentheses:

abx^2 + bcx + adx + cd = 0

and collect the x's:

abx^2 + x(bc + ad) + cd = 0

Now we have to findout what 'a', 'b', 'c', and 'd' are:

abx^2 + x(bc + ad) + cd = 0

x^2 - x - 20 = 0

Let's start at the left. The coefficient (number that is multiplied
by) the x^2 is 1 inour equation. So:

ab = 1

The easiest pair of numbers that works for 'a' and 'b' in the above
equation is when they are both 1. So we can rewrite our identity:

x^2 + x(c + d) + cd = 0Now if we look at the coefficients of x and the constant at the end,
we should be able to get another two equations:

Coefficient of x: -1 = c + d

Constant: -20 = cdYou can either solve those simultaneously, or just look at them.
-5 + 4 = -1 and -5 * 4 = -20. So 'c' is -5, and 'd' is 4.

Usually, you probably wouldn't do all that. When you get used to...