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The quadratic equation in its general form is a second degree equation in one variable and is written as:

ax² + bx + c = 0,

where x is the variable and a, b, and c are constants. For example, if a is 2, b is -3, and c is 5, we get the quadratic equation:

2x² -3x + 5 = 0.

(If a is zero, the quadratic equation reduces to a linear equation.)

All quadratic equations have two solutions, but only if the range of the variable is the extended set of real numbers, to include imaginary numbers

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Factors are numbers being multiplied together, and the factors of any number are the numbers you can multiply together to get the number. So factoring is a process of splitting out the factors of a number that multiply together to give the number.

Take the number 6, for instance. This number has the factors 2 and 3 -- you get the number 6 if you multiply 2 by 3 (or 3 by 2). And the number 4 has the repeat factors, 2 and 2, which says 2 times 2 equals 4. But the number 3 doesn't have any factors, except 3 and 1, which are prime numbers, which aren't solutions of quadratic equations.

It may not seem that an expression like ax² + bx + c can have factors, but remember, it's just a number! Say, for instance, it has the factors (x + d) and (x + e). We then have:

(x + d)(x + e) = 0.

Our knowledge about zero tells us that, in order to get zero, one or the other or both of the two factors has to be zero. If, for instance, (x + d) isn't zero, then (x + e) is. That is,

x + e = 0,

so

x = - e.

Similarly, if (x + e) isn't zero, then (x + d) has to be, and we get:

x + d = 0

so

x = -d.

So (x + d) and (x + e) are the two factors and therefore -d and -e are the two values of x that solve our quadratic equation. 

Factors of an algebraic expression aren't always easy to find. It takes a lot of practice just multiplying factors together, let alone finding them. So let's consider an example.

Say the factors are:

2x + 3

and

x - 2.

To multiply them, we first multiply:

x(2x + 3)

to get

2x² + 3x.

Then we multiply

-2(2x + 3)

to get

-4x - 6.

Adding the two products together gives us

2x² + 3x - 4x - 6

or

2x² -x - 6.

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Sometimes it's not possible to find factors to solve quadratic equations, so you create them. You mustn't change the equation, of course. But you can make apparent changes without changing anything but the form of the equation. You do it by adding 0. The idea is to convert an algebraic expression into a perfect square, or two equal factors.

In fact, you add zero to an algebraic equation by adding a non-zero number to both sides of the equation. You might add 8 to both sides, or d, or d/2, and so on. This would be the same as adding zero to either side, because the respective pairs would cancel.

To solve the quadratic equation

x² - 2x - 4 = 0,

which I drew from College Algebra, by Ohmer and Aucoin, and which can't be factored as it stands, the key is to recognize that the expression x² - 2x + 1 forms a perfect square! In other words, you practically have to know the answer before you can find it.

In any case,

x² - 2x + 1

is the same as

(x - 1)(x - 1).

Multiply the two factors and you will see for yourself.

To form this expression in the equation, now, you have to add 1 to both sides, but you also have to add 4 to both sides. As a result, you change the form to:

x² - 2x + 1 - 4 + 4 = 1 + 4

or

x² - 2x + 1 = 5.

Since the left side is a perfect square, we can take the square root of both sides and get:

(x - 1) = +Ö 5

or

x - 1 = -Ö 5,

because the square root of 5.can be either positive or negative:

In other words, we have the two solutions of our quadratic equation, namely:

x = 1 + Ö 5 and x = 1 - Ö 5.

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The method of completing the square can be used to generate a formula, called the quadratic formula, that can be used to solve any quadratic equation. The approach is to complete the square on  ax² + bx + c = 0.

First we subtract c from both sides and then divide both sides by a, keeping in mind that a isn't zero -- the equation is quadratic, remember. The result is:

(a/a)x² + (b/a)x + (c - c)/a = (0.- c)/a,

which reduces to:

x² + (b/a)x = -c/a.

We can now complete the square by adding (b/2a)² to both sides:

x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²

The left side is now the square (x + b/2a)² and the right side is -c/a + b²/4a².

If we rewrite the right side, we get:

(b² - 4ac)/4a² .

Now we can take the square root of both sides to get the form:

(x + b/2a) = Ö (b² - 4ac)/2a .

Solving for x, we get:

x = -b/2a +Ö (b² - 4ac)/2a .

or

x = -b/2a -Ö (b² - 4ac)/2a.

Combining terms, the quadratic formula is then:

x = (-b +Ö (b² - 4ac))/2a

or

x = (-b -Ö (b² - 4ac))/2a.

Given the specific quadratic equation to be solved -- quadratic meaning that the coefficient, a, isn't zero -- you now merely plug the values of its constants into the quadratic formula. The one condition is that b² - 4ac isn't negative.

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