Thus algebra is full of expressions such as
Note: if the value you give to x is too large then you won't be able to see all of the answer unless you use the arrow keys.
Note: For some calculations involving decimals the answer may
be "slightly off" -- this is a result of the way in which the computer
handles decimals. For a simple example of this take
x = 5.3.
The formula A = r^{2 }expresses the area, A, of a circle in terms of its radius r. That is, given the radius of a circle we can use the formula to find its area. For example, a circle with a three foot radius has an area A = 3^{2 }= 33 = 9, so, taking 3.14 as our approximation for , we get A = 3.149 = 28.26 as the area of the circle. What we have done is replace the "variable" r with the specific number, 3, and the constant symbol, , with its (approximate) numerical value, and evaluated the expression to get a corresponding numerical value for the variable A, that is, the area of the circle..
But what is our problem is that we know the area and we want to find the radius? For example, if a gardener has enough seed to plant 50 square feet of some flower and he wants to plant it in a circular bed then how big a flower bed should he dig? In this case we can use the same formula, A = r^{2}, but this time we know the area, A, and we are trying to find the radius, r. That is, we know that A = 50 = r^{2}. Again, approximating by 3.14, we can rewrite this as 50 = 3.14r^{2}. Dividing both sides by 3.14 we get 15.9236 = r^{2}, then, taking the square root of both sides gives 3.99 = r. Thus, from the practical point of view (of the gardener) the desired flower bed should have a radius of four feet!
We can get a general formula for the desired result
from the formula A
= r^{2
}by
dividing both sides of the equation by
, getting A/
= r^{2} and then taking the square
root of both sides getting the general formula r
=A/,
for the radius when given the area..
Note that this formula says to do just what we did before. That is,
solve the gardeners problem we take A = 50,
divide it by and take the square
root. Of course you may be wondering, just how did we
know what to do? Well, that is one thing you will learn by studying
High School Algebra.
Modern algebraic notation offers many advantages
over earlier notations. Primary among these is that it makes
it easier to manipulate algebraic expressions and thus makes it easier
to explain and apply algebraic concepts.
The use of symbols, equality signs and superscripts is actually a relatively recent development in algebra . Through most of its history algebra was written out using just words and numbers.
The "modern notation" makes it much easier to write down algebraic ideas and results. More important, the use of appropriate notation makes it much easier to perform many of the routine manipulations needed to solve algebraic problems.
Without the use of modern notation it would probably be extremely difficult to write a computer program for the polynomial calculator.
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Algebraic notation has a variety of uses. For example, we can use it to write formulas such as the formula A = r^{2 }for the area of a circle, to give mathematical definitions such as (n/p + r/s) = (ns + pr)/sr for the addition of fractions , to present general mathematical results such as m+n = n+m expressing the commutativity of addition (the fact that the result of adding two numbers does not depend on the order in which we add them), and to describe many kinds of functions. Thus algebraic notation provides a concise way to express many kinds of relationships.
However, even more important is the fact that algorithm notation can
be used to manipulate such expressions so that we can solve a variety of
problems. Much of the material in high school algebra is concerned
with various ways to manipulate such expressions and with the different
kinds of problems we can solve with such manipulations.