Equations 1 

Directions:  Use the computer program I See Graphs to complete the following tasks.

 Click the Standard Equations tab.

 For equation #1, type 2x + 1, and for equation #2, type x2.  Click the yellow Show Tables button.  Three columns of numbers appear in a new window.  the X column represents values of x.  Column 1 represents values of  2x + 1.  Column 2 represents x2.

 Compare the numbers in column 1 with the corresponding numbers in column 2.  At or near the top of the lists, the numbers in column 1 are greater than the numbers in column 2.  After the third number in each list, numbers in column 1 become less than the numbers in column 2.  Is there a value of x for which the numbers in the two columns would be equal?

 To answer that question, we need to look at values of x between integers.  We need to "zoom in" on what is going on.  To do that, we will change the value of DTable.  By default, the value of DTable is 1.  If we make that value smaller, we will be able to compare columns when the value of x is not necessarily an integer.  We also notice that the columns change in relative values somewhere between 2 and 3.

 Begin by double-clicking the 0 beside the Table Minimum button.  Type 2 as the new Table Minimum.

 Double-click the number 1 beside the DTable button.  Type .1 as the new DTable. 

 The table should have been reconstructed using the new values.

 Now as you compare values in column 1 to values in column 2, you should notice that column 2 values become larger than column 1 values between x = 2.4 and x = 2.5.  Now we need to re-define the Table Minimum to be 2.4.   Follow the directions above to change 2 to 2.4.

 Double-click the number .1 beside the DTable button.  Type .01 as the new DTable. 

 Now we see that the two column's values "cross" between 2.41 and 2.42, so change Table Minimum to 2.41, and change DTable to .001.  The columns now cross between

x = 2.414 and x = 2.415.

 Change Table Minimum to the smaller of these two numbers, 2.414, and change DTable to .0001.  Now we see that column 1 and column 2 are equal to 4 decimal places (5.8284) when x = 2.4142.  In other words, 2x + 1 = x2 when x = 2.4142.  Of course, this is just a decimal approximation to the actual solution to this equation.  The actual solution is an irrational number, i.e., it has a non-repeating, non-terminating decimal representation.

Because this is a quadratic equation (contains x2), we should expect another solution.

Change Table Minimum back to 0, and change DTable back to 1.

Click the up-arrow to scroll the columns the other direction from before.

 With just a couple of clicks, values in column 1 become greater than values in column 2.

 Adjust Table Minimum and DTable as we did above and find a second solution to the equation 2x + 1 = x2.

 1)  What is the other solution to the nearest 1/10000?  ___________________

 Use a similar technique to find solutions to the following equations:

 2)  3x - 1 = 2x2

 3) 

 4)   x3 - 5x2 + x - 4 = 2x - 6

 5)  x3 - 4x2 + 3x + 2 = 0

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