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Suppose you need to simplify the square root of a number. The easiest way to do this is to break the number under the radical sign into factos, one of which is a perfect square.  For example,

square root(84)

Get factors of 84, one of which is a perect square.  We could choose 2 and 42, but neither of those is a perfect square, so it's better to choose 4 and 21.

Therefore,

square root(84) = square root(4) * square root(21)

We know the square root of 4 is 2, therefore

square root(84) = 2square root(21).

Since the only factors of 21 are 1, 3, 7 and 21, all of which are not perfect squares,
square root(21) cannot be simplified.

The book is now available in paperback. Check out the link below.

http://www.lulu.com/spotlight/KKauffman1969

The new book is available as an ebook for your pc. Not formatted for Nook or Kindle, sorry. But the paperback should be out soon.
http://www.lulu.com/shop/kerry-kauffman/algebra-simplified-intermediate-advanced/ebook/product-20093947.html

Front cover for next book, which will be out shortly.  I plan on working on a statistics book next.

Suppose we have the following problem to solve:


3^(2x + 1) = 27^(3x)

We can solve this by getting a common base and setting the exponents equal to each other. We know that 3^3 = 27, therefore..


3^(2x + 1) = 3^(9x)

Now that the bases are the same (3), we set the exponents equal to each other and solve for x.

2x + 1 = 9x

1 = 7x

1/7 = x


We can also solve this problem by taking the log of both sides and using rules for logarithms.

log(3^(2x + 1)) = log27^3x

(2x + 1)log3 = (3x)log27

(2x + 1)/3x = log27/log3

(2x + 1)/3x = 3

9x = 2x + 1

1 = 7x

1/7 = x

In this problem is was simpler to get a common base. Sometimes it's quite difficult to get a common base. In those cases it's easier to take the log of both sides first.

When using matrices to solve systems of equations, remember we can interchange rows, add or subtract rows to form new rows, multiply rows and add to other rows or divide rows and add to other rows. All of this is in an attempt to get the matrix into a form where we can solve for the variables in the equations of the system. One form is seen above.  Suppose the variables in the system are x, y, z. Then z = f. Substitute f for z in the second equation to solve for y and then substitute that value for y into the first equation and solve for x.