Solve a Quadratic Equation by using the Square Root Method

This Movie (Quicktime)
will show you how to solve quadratic equations by using the Square Root Method.

The following is the movie transcript.

Hi, my name is Anthony and I am going to show you how to solve quadratic equations using the square root method.

Here is a quadratic equation that can be solved using the square root method.
Seven ‘x’ squared plus three equals eight hundred and fifty.

The first step is to apply the addition property, which means we need to transpose positive three to the opposite side of the equation which will give us minus three.
This will give us seven ‘x’ squared equals eight hundred and fifty minus three.
Which is seven ‘x’ squared equals eight hundred and forty seven.

The next step is to divide by the coefficient of ‘x’ squared, which is seven.
So, you take seven ‘x’ squared and divide by seven, you take eight hundred and forty seven and divide that by seven also, which gives us ‘x’ squared equals one hundred and twenty one.

For the final step we will apply the square root method, which means we need to take the square root of x’ squared plus or minus the square root of one hundred and twenty one which will give us ‘x’ equals plus or minus eleven.

This concludes my presentation on how to solve a quadratic equation using the square root method.

End Movie Transcript

Posted at 01:40PM Feb 24, 2007 by llinebarger in Intermediate Algebra  | 

Finding Slope and Y-Intercept

This Movie (Quicktime)
will show you how to find the slope and y-intercept given the equation of a line.

The following is the movie transcript.

Hi my name is Paul and my topic is Slope and the slope-intercept form of equations. I’m going to show you how to find the slope and y-intercept given the equation of a line and find the equation of a line given the slope and y-intercept.

When an equation is written in the form y = mx + b, the slope of the line and y-intercept of a line can be determined by inspection.  Here we'll find the slope and y-intercept given the equation of a line y = mx + b where m is the slope and b is the y coordinate of the y-intercept.

For example, we’ll find the slope and y-intercept of the lines with the following equations. In this equation y equals three-fifths x plus 2, the slope is three over five and the y-intercept is 2 or the ordered pair (0,2). Another example, y equals 3x minus 4, the slope equals 3 or 3 over 1 and the y-intercept equals negative 4 or the ordered pair (0, negative 4).

In this last example, which is written in standard form, we have 2x minus 5y equals 10. We’ll solve for y, negative 5y equals negative 2x plus 10. Divide by negative 5 on both sides of the equation and you end with y equals two-fifths x minus 2. Your slope will equal two-fifths and your y-intercept will equal negative 2 or the ordered pair (0,negative 2).

For the second part of my topic, we want to find the equation of a line given the slope and y-intercept.
First, use the slope-intercept form of an equation of a straight line y = mx + b.
Second,you want to substitute known values for mx+b.
Finally, you write the equation in standard form or slope-intercept form.

For example, write the equation for a line with a slope of negative 3 and a y-intercept of 5. We know the slope is negative 3 and the y-intercept is 5. So substitute the known values in the equation and you come up with y equals negative 3x + 5.

This is my podcast on slope and the slope-intercept form of equations.

End Movie Transcript

This podcast was created by Paul in Mrs. Linebarger's Algebra class at Southwest Tennessee Community College.  www.southwest.tn.edu.

http://ww4.southwest.tn.edu/podcast/llinebarger/Paul.m4v

Posted at 01:10PM Feb 09, 2007 by llinebarger in Intermediate Algebra  | 

Quadratic Formula

This Movie (Quicktime)
will show you how to solve a quadratic equation using the quadratic formula.

The following is the movie transcript.

Hi my name is Shelby and today I’m going to show you how to solve a quadratic equation by using the quadratic formula.  By using the quadratic formula, you can always find the solution for ax to the second power plus bx plus c equals zero.  The quadratic formula is simply x equals the opposite of b plus or minus the square root of b to the second power minus four times ac over two a. Note that in order for the formula to work you must have the quadratic equation equal zero.

The problem that we are going to solve today will be x to the second power plus three x equals four. Setting this equation equal to zero will give you x to the second plus three x minus four equals zero.

By substituting the numbers from the problem into the formula a will become one b will become three and c will become negative four giving us the problem x equals a negative three plus and minus the square root of three to the second power minus four times one time a negative four all over two times one.

By using the order of operations you will solve the problem which will be x equals a negative three plus and minus the square root of nine plus sixteen over two. X equals a negative three plus and minus the square root of twenty five over two, leaving us to take the square root of twenty five.

By plus and minus being between the negative three and the five you must add and subtract the problem, giving you negative three minus five over two equals a negative eight over 2 leaving your first answer a negative four. The next one will be a negative three plus five over two equals two over two leaving you last answer a one.

This concludes how to solve a quadratic equation by using the quadratic formula.

End Movie Transcript

Posted at 06:20PM Dec 10, 2006 by llinebarger in Intermediate Algebra  | 

Dividing Rational Expressions

This Movie (Quicktime)
will show you how to divide rational expressions.

The following is the movie transcript.

Hi this is Raphael.  Today we are going to divide rational expressions. The problem we are going to do is y plus five over y squared minus four y minus five divided by y squared plus four y minus five over y plus one.

For our first step in dividing rational expressions, you will need to rewrite the problem as the multiplication of the reciprocal of the second term. This will give us y plus five over y squared minus four y minus five times y plus one over y squared plus four y minus five.

For your second step in dividing rational expressions, you will need to factor the numerator and the denominator this will give us y plus five over the quantity y minus five times the quantity y plus one times y plus one over the quantity y plus five times the quantity y minus one.

For your third step in dividing rational expressions, you will need to cancel out the like terms. The y plus five in the numerator and denominator cancel each other out and y plus one in the numerator and the denominator cancel each other out. Finally if there is nothing in the numerator you will need to put a one to hold its place, leaving you with one over the quantity y minus five times one over the quantity y minus one in the denominator.

For your final answer, you will get one over the quantity y minus five times the quantity y minus one.

This concludes my presentation of dividing rational expressions.

End Movie Transcript

Posted at 06:12PM Dec 10, 2006 by llinebarger in Intermediate Algebra  | 

Graphing Interval Expressions and Writing Interval Notation

This Movie (Quicktime)
will show you how to graph interval expressions and write the interval notation.

The following is the movie transcript.

Hello this is Ashley and today I am going to show you how to graph an inequality on a number line and write it in Interval Notation.

There are three rules to graphing an inequality.

First is to determine the boundaries of the inequality if they exist.

The second rule is to locate the boundaries on a number line. When the inequality is less than or greater than, the boundaries are not included. This is represented with parenthesis. You would use a closed parenthesis for less than and an open parenthesis for greater than. If the inequality is less than or equal to or greater than or equal to the boundaries are included. You would represent this with brackets. You would use a closed bracket for less than or equal to and an open bracket for greater than or equal to.

The third rule would be to shade the portion of the number line between the boundaries. If there is no boundary in one or either direction the graph continues indefinitely in that direction. In this case you would use an infinity sign.  This movie will go over how to graph these nine inequalities and write them in Interval Notation.

For our first example we will graph x is less than three. First you would start by writing a closed parenthesis over your three and shade to the left because the number of x is less than three and you would draw an arrow to indicate that it goes on infinity in that direction. To write your Interval Notation you would use an open parenthesis and it would be negative infinity and stops at three and three is not included so you use a closed parenthesis. 

Our next example we will use x is greater than negative two. Again you would start by using an open parenthesis over negative two and you would shade to the right to indicate x is greater than negative two.  For the Interval Notation you would use open parenthesis negative two and an infinity sign closed parenthesis. 

In our next example we will use x is less than or equal to two. In this example we would use a bracket because x is less than or equal to two. You would use a closed bracket to indicate two is include, and shade to the left.  The Interval Notation for this would be open parenthesis negative infinity and two and then a closed bracket to indicate two is included.

For our next example we will use x is greater than or equal to negative one. You would start by using an open bracket over negative one to indicate negative one is included and shade to the right.  And the Interval Notation for this would be open brackets negative one, infinity sign closed parenthesis. 

Our next example will be negative one is less than x and x is less than three. You would start by using an open parenthesis over negative one and a closed parenthesis over your three and x will equal everything in between.  The Interval Notation would be open parenthesis negative one and three closed parenthesis.

Our next example we will use negative one is less than or equal to x and x is less than or equal to three. In this case you would use brackets to indicate negative one and three are included and x will equal everything in between. Your Interval Notation for this would be open bracket negative one and three then closed bracket. This indicates negative one and 3 are included.

In this example one is less than x and x is less than or equal to four. So you would use an open parenthesis over your one and a closed bracket over your four and x would equal everything in between. And the Interval Notation would be open parenthesis one and four then a bracket indicating four is included.

Next we will use negative one is less than or equal to x and x is less than three. Again we will start by using a bracket for negative one and a closed parentheses for the three and x will equal everything in between. And for your Interval Notation you will use a bracket negative one and three closed parentheses.

For our last example we will use for the set of all x such that x is equal to all real numbers. This means all numbers are included. So you would shade the entire number line. The Interval Notation for this would be open parenthesis negative infinity and it goes on infinity with a closed parenthesis.

This will conclude my quick time movie on graphing inequalities and writing them in Interval Notation.

End Movie Transcript

Posted at 11:33AM Dec 05, 2006 by llinebarger in Intermediate Algebra  | 

Subtracting Rational Expressions

This Movie (Quicktime)
will show you how to subtract rational expressions with unlike denominators.

The following is the movie transcript.

Hi, my name is Therese and this podcast presentation is about Subtracting Rational Expression with unlike Denominators.

To subtraction a rational expression with unlike denominators you must follow these four steps.
• Find a common denominator
• Change each fraction to an equivalent fraction
• Subtract the numerators while keeping the same denominator
• Reduce if possible

The problem we are going to subtract is five over x minus three over four x. To subtract this rational expression with unlike denominators, you must first establish the common denominator, which in this case is four x. After identifying the common denominator you should change each fraction to an equivalent fraction with the common denominator of four x.
To change the first fraction to the common denominator we will multiply the numerator by four and the denominator by four. Since the second fraction already has the common denominator of four x we don’t need to change it.
Now we will multiply the fraction: four times five gives us twenty and four x gives us four x, subtracting three over four x. Next we will subtract the numerator while keeping the same denominator twenty minus three gives us seventeen over four x. And since we can not reduce the final answer is seventeen over four x.
This concludes the presentation on how to subtract rational expressions with unlike denominator.

End Movie Transcript

Posted at 11:26AM Dec 05, 2006 by llinebarger in Intermediate Algebra  | 

Graphing with Slope and Y-intercept

This Movie (Quicktime)
will show you how to graph a linear equation using slope and y-intercept.

The following is the movie transcript.

Hello, this is Cari and I am going to show you how to find slope and y-intercept to graph a linear function.

Rewrite the following equation in slope–intercept form. Two x minus three y equals nine. Negative three y equals negative two x plus nine. Divide by negative three so y equals two-thirds x minus three.

To graph a linear equation by using slope and y-intercept method y equals mx plus b. There are four steps in graphing a line with this method. One, locate the y-intercept on the y-axis. Two, using the slope, determine the amount of vertical movement. Three, locate additional points on the graph by counting the horizontal and vertical movements. Four, draw a line through the points to connect.

For our problem, we are going to locate the y-intercept on the y-axis which is negative three. Step two, using the slope; determine the amount of vertical and horizontal movement. Two indicates vertical movement and three indicates horizontal movement. Step three, locate additional points on the graph by counting up and over. Step four, draw a line through the points to connect.

This concludes how to find slope and y-intercept to graph a linear equation.

End Movie Transcript

Posted at 01:43PM Nov 17, 2006 by llinebarger in Intermediate Algebra  | 

Solving Quadratic Equations by Factoring

This Movie (Quicktime)
will show you how to solve Quadratic Equations by Factoring.

The following is the movie transcript.

Hello. My name is Natasha and my podcast presentation is on Solving Quadratic Equations by Factoring.

To solve an Incomplete Quadratic Equation by Factoring : set the equation equal to zero, factor the equation, set the factors equal to zero, solve each equation. One root will always be zero.

For example, if you have x squared minus eight x equals zero, you should identify this equation as an incomplete quadratic equation. It doesn’t have a constant term. This equation is already set equal to zero, so the next step is to factor the equation. In this case, the common factor is x so your result after factoring should be x times the quantity of x minus eight equals zero. Next, you should set each factor equal to zero. X equals zero, and x minus eight equals zero. Solve for x by adding eight to each side. X minus eight plus eight equals zero plus eight. X should equal eight. So your two solutions are x equals zero and x equals eight.

To solve a Complete Quadratic Equation by factoring: arrange the equation in standard form, factor the equation, set each equation equal to zero, solve each equation. The roots will usually be different.

Let’s take a look at the next example. X squared plus six x equals negative five.
You should identify this equation as a complete quadratic equation. First, I will arrange the equation in standard form which is x squared plus six x plus five equals zero. Next, factor the equation. There are no common factors so you must find two numbers that you can multiply and get five, but add to get six. After factoring, you should come up with x plus five times x plus one equals zero. Set the factors equal to zero and solve.
X plus five equals zero. X plus five minus five equals zero minus five. X should equal negative five. X plus one equals zero. X plus one minus one equals zero minus one. X equals negative one. So your two solutions are x equals negative five and x equals negative one.

This concludes how to solve Quadratic Equations by Factoring.

End Movie Transcript

Posted at 06:35PM Nov 05, 2006 by llinebarger in Intermediate Algebra  | 

Adding Rational Expressions with Unlike Denominators

This Movie (Quicktime)
will show you how to add Rational Expressions with Unlike Denominators.

The following is the movie transcript.

Hello. My name is Lindsay, and in this QuickTime movie I will explain how to add Rational Expressions with Unlike Denominators.

To add Rational Expressions with Unlike Denominators, five steps must be followed:
Find the common denominator, change each fraction to an equivalent fraction with the common denominator, add the numerators, keep the same denominator, reduce if possible.

I will work the following problem for you: four over three X plus five over six.
First, convert to equivalent fractions with a common denominator. Eighteen X is the common denominator on this problem. For four over three X, we need to multiply the numerator and the denominator by six. For five over six, we need to multiply the numerator and denominator by three X.
Six times four gives us twenty-four, and three X times six gives us eighteen X, plus five times three X gives us fifteen X, and three X times six gives us eighteen X.
Now that the denominators are alike, add the numerators and keep the common denominator. This will give us twenty-four plus fifteen X, all over eighteen X. Combine like terms in numerator if possible. There are no like terms to combine in this problem.
Last, reduce. In this case, all numbers are divisible by three. So, three goes into eighteen X six times, so it becomes six X. Three goes into twenty-four eight times, so that becomes eight and fifteen X, three goes into it five times, so it becomes five X. Now, rewrite the problem. The problem can no longer be reduced so the final answer is eight plus five X over six X. This concludes how to add rational expressions.

End Movie Transcript

Posted at 12:13PM Oct 27, 2006 by llinebarger in Intermediate Algebra  | 

Graph a Linear Function with x- and y-intercepts

This Movie (QuickTime)
will show you how to graph a linear function by using the x- and y-intercepts.

The following is the movie transcript.

My Podcast Presentation is on "Finding Intercepts and Graphing a Linear Function".

For example, if you have y= 2x + 4, you can find the x and y intercepts, which gives you the linear function. To find the x-intercept, you set y = 0. So you have 2x + 4= 0, you subtract 4 from the left side and 4 from the right side. Then you will end up with 2x = negative 4. And then to solve for x, you divide both sides by 2. So x= negative 2.  So now you solve for your x-intercept. Your point will be (negative 2, 0).

Now to find the y-intercept, you have to set x=0. So you have y= 2(0) + 4. So then you will have y = 0+4. So then you will have after you add 4 + 0, y = 4. So your point for the y-intercept is at (0, 4).

Now you have to graph the two points. Since for the x-intercept we have (negative 2, 0), we have to plot that on the x-axis. So we start at the origin and go over negative 1, negative 2 and plot a point, since the y is 0. Now for the y-intercept, we have (0, 4). So we start at the origin, and we plot (0, 4) on the y-axis. So we go up four spaces. Now we’re ready to draw our line for the linear function. Now all I need to do is draw a line through the two points to have my linear equation graphed.

This concludes how to find intercepts and graph a linear function.

End Movie Transcript

Posted at 11:16AM Oct 18, 2006 by llinebarger in Intermediate Algebra  | 

What Is Slope?

This Quick Time Movie
will show you how the slope of a line is defined and graphically show what we mean by the slope of a line.

The following is the movie transcript.

What Is Slope?

This Quick Time movie will show how the slope of a line is defined and graphically show what we mean by the slope of a line.

The slope of a nonvertical line measures a line’s steepness.

As we move along this line from left to right, from one point to another, we measure the steepness by comparing the vertical change (how far we have moved up) to the horizontal change (how far we have moved to the right).

Slope is usually denoted with the letter m.
The slope of a line is the ratio of the change in y over the change in x. This is also known as the rise over the run.

Using two points a line passes through, we can find this ratio. To find the slope of this line we need to pick two points this line passes through.

For the first point, I will choose (negative two, negative five) and for the second point I will choose (four, four).

I can graphically find the rise and run by first constructing lines to show the vertical change (how far I move up) and the horizontal change (how far I move to the right) to get from my first point to my second point.
Next, I will count the length of my constructed lines.
Counting along the rise, I get nine units.
Counting along the run, I get six units.

Now I can write the slope m equals rise over run which equals nine over six. We can reduce this to three over two.

Note that as I count up three units and to the right two units, I end up on the line. Again, if I move up three units and to the right two units, I end up on the line.

This indicates that I can choose any two points on a line to obtain its slope.

We can also find the slope of a line without using a graph.

The lesson ‘Finding Slope Using Two Points’ will show how to find the slope of a line given two points that the line passes through.

This concludes the lesson ‘What is Slope?’

Posted at 03:14PM Jul 27, 2006 by llinebarger in Intermediate Algebra  | 

Graph The Linear Equation: 2x-y=4

This Quick Time Movie
will show you how to graph the linear equation
2x-y=4
by finding three ordered pair solutions.

The following is the movie transcript.

Graph the linear equation two times x minus y equals four.
Let’s first look at the steps we should follow to graph a linear equation.

To graph a linear equation:

• First, find at least three ordered pair solutions.
  Order pairs can be graphed as points in a rectangular coordinate system. Two points will determine a line, but the third point will insure your graph and solutions are correct.
• Second, choose your pairs so the points are not too close together. This will result in a more accurate graph.
• Third, graph the ordered pairs and then
• Fourth, draw a line through the plotted points. Place arrows at each end of the line to indicate the line continues on in both directions.

Starting with Step 1 we need to find three ordered pair solutions.

The solution to a linear equation in two variables requires two numbers, one number for x and one number for y. When we replace the variables in our equation with these two numbers we want to end up with a true statement.

These two numbers form an ordered pair, because we write them in parentheses with the number for x first and then the number for y.

A linear equation has an infinite number of ordered pair solutions, we need to find only three pairs to graph the line.

To find three ordered pairs, we need to choose numbers that will fit in our rectangular coordinate system and we need to choose values that are not too close together.

This is the rectangular coordinate system I am going to use to graph our ordered pairs. It goes from negative six to positive six on both the x-axis and the y-axis. So I will need to choose points that have x and y values that are between negative six and positive six.

For the first ordered pair, I am going to let x equal negative one. Substituting negative one into the equation we get two times negative one minus y equals four.
Simplifying the left side we get negative two minus y equals four.

Transposing negative two to the opposite side gives us four plus two and leaves us with the opposite of y equals six which we will change to y equals negative six.

This gives us our first ordered pair negative one, negative six.
We check our solution by substituting the ordered pair into our equation, negative one in for x and negative six in for y. This gives us two times negative one minus negative six equals four. Simplifying the left side, we get negative two plus six equals four. On the left side, negative two plus six equals four so we are left with four equals four which is a true statement. This means negative one, negative six is one solution to our equation.

We now have to find two more solutions before we graph our line.

This screen shot shows finding our second ordered pair two, zero. To read through the solution pause the movie now. Or you can go to the text version for this movie to read the solution.

Text for finding the second ordered pair two, zero:
Let x equal two. Substituting two into the equation we get
Two times two minus y equals four.
Simplifying the left side we get four minus y equals four.

Transposing the four to the opposite side gives us four plus negative four which is zero and leaves us with the opposite of y equals zero which we change to y equals zero.

This gives us our second ordered pair two, zero.

This screen shot shows finding our third ordered pair four, four. To read through the solution pause the movie now. Or you can go to the text version for this movie to read the solution.

Text for finding the third ordered pair four, four:
Let x equal four. Substituting four into the equation we get
Two times four minus y equals four.
Simplifying the left side we get eight minus y equals four.

Transposing the eight to the opposite side gives us four plus negative eight which is negative four and leaves us with the opposite of y equals negative four which we change to y equals four.

This gives us our second ordered pair four, four.

Now that we have our three ordered pairs (negative one, negative 6), (two, zero), and (four, four) we are ready to graph them in our rectangular coordinate system.

To graph the ordered pair (negative one, negative six), I will start at the origin and move one unit to the left on the x-axis and then move down six units and place a point.

To graph the ordered pair (two, zero), I will start at the origin and move two units to the right on the x-axis and place a point. The zero for y indicates no up or down movement.

To graph the ordered pair (four, four), I will start at the origin and move four units to the right on the x-axis and then move up four units and place a point.

Now we are ready to draw a line through our three points and place arrows at each end of our line to indicate it goes on in both directions.

This concludes the solution to our problem.

Posted at 10:18PM Jun 26, 2006 by llinebarger in Intermediate Algebra  | 

Solve a Quadratic Equation by the Square Root Method

Solve a Quadratic Equation by the Square Root Method

Example 1

Solve and round to the nearest thousandths, nine x squared plus five equals seven. This is a quadratic equation. When we solve an equation we want to find all values of the variable that will make the equation true. To solve an equation we need to isolate the variable. To do this for our problem we will first subtract five from each side of the equation. Here I have copied our equation and I have shown that I am going to subtract five from the left side of the equation and I am also going to subtract five from the right side of the equation. When we are solving an equation, whatever we do to one side of the equation we must also do the same thing to the opposite side. This means we have to use the same operation with the same number to both sides of the equation. Subtracting five from both sides, leaves us with nine x squared equals two. Since the coefficient of the x squared term is being multiplied by nine, we will need to divide each side of this equation with nine to undo the multiplication. Dividing the left side with nine, will leave us with x squared and dividing the right side by nine will leave us with two over nine or two-ninths. Now we will use the Square Root Property that says if our squared variable is equal to a positive number then we can take the square root of both the left side and the right side. The Square Root Property also says that if we take the square root of x squared, it will equal both the positive square root of two-ninths and the negative square root of two-ninths. Now let’s write this so we can see that we actually have two solutions. We will write this as x equals the square root of two-ninths and x equals the negative square root of two-ninths. We are not quite finished because we can simplify our two fractions. In the denominator, we have nine which is a perfect square and we can take the square root of nine and get three. Our answer then is going to be x equals the square root of two over three and x equals negative square root of two over three. We have one more step to do to finish this problem because the directions say to round this answer to the nearest thousandths. If we leave the answer in the form the square root of two over positive three or negative square root of two over three, we call these exact answers. The square root of two is an irrational number and we will not be able to list it in an exact decimal form so we will have to round it. Since the directions say to round to the nearest thousandths, we will want to write our answer with just three decimal places after the decimal point. Using a calculator, we take the square root of two, then divide by three, and then round to three decimal places after the decimal point. This will give us x equals zero point four seven one or we can get x equals negative zero point four seven one.

Quick Time Movie

Posted at 05:26PM Jun 14, 2006 by llinebarger in Intermediate Algebra  | 

Linear Inequalities - Solve: -6x-3<21

Solve the linear inequality negative six x minus three is less than twenty-one. Graph the solution set and write the solution set using set notation and interval notation.  Solving a linear inequality is much like solving an equation.  We are going to want to isolate the variable on the left side of this inequality.  To do that we are going to start by using the addition property and add positive three to each side of this inequality.  Adding three to each side of this inequality will leave us with negative six x is less than twenty-four. Since the coefficient of our x term is a negative number, that means we are going to need to divide both sides of this inequality with negative six.  Remember that the property for division with a negative number on inequalities requires us to change the direction of the inequality sign. As an example, let's review this property for just a minute.  If we have eight is less than ten, and we divide each side by negative two without changing the direction of the inequality sign, this is going to leave us with negative four is less than negative five.  And of course, this is a false statement. So when we divide with negative six here, we are going to need to change the direction of this inequality sign. Dividing each side of this inequality now with negative six, on the left side is going to give us x and on the right side side will leave us with negative four. So our solution is x is greater than negative four. Now we need to graph this solution set, so let me insert a number line.   Now that we have our number line, we can graph the solution set for the linear inequality.  Our solution, x is greater than negative four, means that our solution will include all numbers up to and greater than negative four, but will not include negative four.  We show this on a graph by putting a parentheses at negative four and drawing an arrow to the right.  The parentheses at negative four indicates that negative four is not part of the graph. Now we are ready to do the last part of this problem which is to write the solution set using set notation and interval notation.  For set notation, we are going to write the set of all x such that x is greater than negative four.  For interval notation, we are going to write using our parentheses the same as we started our graph with negative four comma and then we are going to indicate that this graph goes off to the right to infinity so we use our symbol for infinity and this always closes with parentheses. This concludes the solution to our problem.

Quick Time Movie

Posted at 01:21PM Jun 12, 2006 by llinebarger in Intermediate Algebra  | 

Polynomials-Reduce Fractions 1b

Simplify Rational Expression

Example 1b Text

Let's work this problem a second way now using the Laws of Exponents.   To start with we are going to take twenty seven and eighteen and cancel out the common factor nine the same way we did in the first method.   I will show that by drawing a line through twenty-seven. Twenty-seven divided by 9 will leave us with three and eighteen divided by nine will leave us with two.  For the variable parts we will use the law of exponents that says if the bases are alike and you are dividing, then you can take the exponents and subtract them.  Here we have a common base of a and the exponents are three and two.  So we can write a to the three minus two and this will leave us with a to the first.  Next, we have the factors of b and the exponents here are in the numerator we have b to the first power and in the denominator we have b to the fourth.   So we can write b to the one minus four which equals b to the negative third.  And our last factor which is c squared over c squared we can write as c to the two minus two which equals c to the zero.  Now c to the zero power, if we remember our rules for exponents, any number to the zero power is equal to one.  So actually what this is saying is that if we have two bases with the same exponents that is the same as having a number over itself and a number over itself is always equal to one.  Now lets rewrite this back together as a single fraction.  And we will have three a over two b to the third power.  Remember that our Laws of Exponents say that if we have a negative exponent that we need to write this in the reciprocal form which is one over b to the third power.  So our b is going to end up in the denominator.

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Posted at 11:13AM Jun 08, 2006 by llinebarger in Intermediate Algebra  |