Changes

==Elementary Algebra==

==Elementary Algebra==

While algebra is a very wide-ranging and advanced topic in theoretical mathematics, the word also refers to a very important topic in middle-school and high-school mathematics.  This topic, which is sometimes known as "elementary algebra" to distinguish it from the more advanced branches, is the handling of arithmetical expressions and operations involving quantities that aren't known numbers, but are symbols like x, y, or z.  These symbols are generally referred to as "variables" or "unknowns".

While algebra is a very wide-ranging and advanced topic in theoretical mathematics, the word also refers to a very important topic in middle-school and high-school mathematics.  This topic, which is sometimes known as "elementary algebra" to distinguish it from the more advanced branches, is the handling of arithmetical expressions and operations involving quantities that aren't known numbers, but are symbols like ''x'', ''y'', or ''z''.  These symbols are generally referred to as "variables" or "unknowns".

Whereas, in elementary arithmetic a typical question might be:

Whereas, in elementary arithmetic a typical question might be:

::What is the value of 3+4?

::What is the value of 3+4?

in elementary algebra a typical question might be:

in elementary algebra a typical question might be:

::If x+4=7, what is the value of x?

::If ''x''+4=7, what is the value of ''x''?

Of course this example is so simple as to be practically pointless—why would someone ask a question this way?—the use of algebra opens up sophisticated ways of solving sophisticated problems.

Of course this example is so simple as to be practically pointless—why would someone ask a question this way?—the use of algebra opens up sophisticated ways of solving sophisticated problems.

Before continuing, let's see how we actually solved that.  We subtracted 4 from each side of the equation.  On the left side, subtracting 4 canceled the "+4", so we got "x+4-4", which is, of course, just x.  On the right side, we got 7-4, which is 3.  We used the principle that

Before continuing, let's see how we actually solved that.  We subtracted 4 from each side of the equation.  On the left side, subtracting 4 canceled the "+4", so we got "''x''+4-4", which is, of course, just ''x''.  On the right side, we got 7-4, which is 3.  We used the principle that

::You can apply mathematical operations to something, even if it contains unknowns like "x".

::You can apply mathematical operations to something, even if it contains unknowns like "''x''".

The goal was to get an equation that has just "x" on the left side, with some number on the right.  In the original problem, x was on the left side, but hidden with some other mathematical operations.  Our job was to strip off those operations, one at a time.  Here's a more complicated example:

The goal was to get an equation that has just "''x''" on the left side, with some number on the right.  In the original problem, ''x'' was on the left side, but hidden with some other mathematical operations.  Our job was to strip off those operations, one at a time.  Here's a more complicated example:

::<math>\left(\frac{\sqrt{x+31}}{3} + 1\right) \times 6 = 18\,</math>

::<math>\left(\frac{\sqrt{x+31}}{3} + 1\right) \times 6 = 18\,</math>

We undo the multiplication, by dividing both sides by 6:

We undo the multiplication, by dividing both sides by 6:

so we get

so we get

::<math>(3+4) \times x = 56\,</math>

::<math>(3+4) \times x = 56\,</math>

and hence x=8.  You can plug x=8 into <math>3 \times x + 4 \times x\,</math> and see that the result is 24+32, or 56.

and hence ''x''=8.  You can plug ''x''=8 into <math>3 \times x + 4 \times x\,</math> and see that the result is 24+32, or 56.

==Polynomial Manipulations==

==Polynomial Manipulations==

so

so

::<math>\frac{x^2 - 8 x + 15}{x-3} = x-5\,</math>

::<math>\frac{x^2 - 8 x + 15}{x-3} = x-5\,</math>

from which one gets x=12.

from which one gets ''x''=12.

This was just a case of the principle that you can perform arithmetic operations even when the expression contain unknowns.  In this case we were actually applying the distributive law 3 times.

This was just a case of the principle that you can perform arithmetic operations even when the expression contain unknowns.  In this case we were actually applying the distributive law 3 times.

::<math>(x-3) \times (x-5) = (x) \times (x-5) + (-3) \times (x-5) = (x) \times (x) + (x) \times (-5) + (-3) \times (x) + (-3) \times (-5)\,</math>

::<math>(x-3) \times (x-5) = (x) \times (x-5) + (-3) \times (x-5) = (x) \times (x) + (x) \times (-5) + (-3) \times (x) + (-3) \times (-5)\,</math>

When we multiply the sum of two things times the sum of two things, we get the sum of four things.  This is an extremely common operation, and there is a mnemonic for it&mdash;"FOIL"&mdash;which stands for "first, outside, inside, last".  We multiply the first item in each parenthesized expression; that's x times x.  We multiply the "outside" items; that's x times -5.  The "inside" items are -3 times x, and the "last" items are -3 times -5.

When we multiply the sum of two things times the sum of two things, we get the sum of four things.  This is an extremely common operation, and there is a mnemonic for it&mdash;"FOIL"&mdash;which stands for "first, outside, inside, last".  We multiply the first item in each parenthesized expression; that's ''x'' times ''x''.  We multiply the "outside" items; that's ''x'' times -5.  The "inside" items are -3 times ''x'', and the "last" items are -3 times -5.

This same principle applies when there are more than two items in each factor; it just doesn't have a useful mnemonic word.  Each term in the first factor (first parenthesized expression) get multiplied by each term in the second factor, and all the products are added up.

This same principle applies when there are more than two items in each factor; it just doesn't have a useful mnemonic word.  Each term in the first factor (first parenthesized expression) get multiplied by each term in the second factor, and all the products are added up.

==Quadratic equations==

==Quadratic equations==

A quadratic equation is a type of equation in which the degree of x is exactly 2.  Usually, these equations have two possible values of x.  An example of a quadratic equation would be:

A quadratic equation is a type of equation in which the degree of ''x'' is exactly 2.  Usually, these equations have two possible values of ''x''.  An example of a quadratic equation would be:

::<math>x^2 - 5x + 6 = 0\,</math>

::<math>x^2 - 5x + 6 = 0\,</math>

::<math>x - 3 = 0\,</math>

::<math>x - 3 = 0\,</math>

Thus x must equal either 2 or 3.  There is no way to tell which is the "real" value of x, so the actual answer is written as:

Thus ''x'' must equal either 2 or 3.  There is no way to tell which is the "real" value of ''x'', so the actual answer is written as:

::<math>x = 2, 3\,</math>

::<math>x = 2, 3\,</math>

This is just shorthand for "x is either 2 or 3."

This is just shorthand for "''x'' is either 2 or 3."

===Completing the square===

===Completing the square===

the left-hand side is not a perfect square trinomial.  That is to say, if we tried to take the square root of it, we couldn't simplify it any further, so that wouldn't help us in solving the equation.  We need to change the equation to make it a perfect square.

the left-hand side is not a perfect square trinomial.  That is to say, if we tried to take the square root of it, we couldn't simplify it any further, so that wouldn't help us in solving the equation.  We need to change the equation to make it a perfect square.

The key here is the coefficient of the linear term, which is the number that's multiplied by the x that isn't squared.  For us, that's 5.  We divide that number by 2 and then square it, which yields

The key here is the coefficient of the linear term, which is the number that's multiplied by the ''x'' that isn't squared.  For us, that's 5.  We divide that number by 2 and then square it, which yields

::<math>(5/2)^2 = 25/4 = 6 + 1/4\,</math>

::<math>(5/2)^2 = 25/4 = 6 + 1/4\,</math>