A quadratic equation in standard form is one of the form
ax2+bx +c=0 let us look at the example below:
x2-8x +15=0
Suppose we plot y = x2-8x +15
Notice the equation states x2-8x +15 =0
And y= x2-8x +15 so we can say our answer is the x that occurs on our
graph where y=0. (why?)
So our answers are located where y=0. On a graph y=0
anywhere on the x axis.
The x axis is the very heavy thick horizontal line at
left. Our answer is the x value where the graph crosses the x axis. These points
are the center of the two circles . We can see that the coordinates of these
points are :
X=3, y=0 and x=5,y=0. That means that when x=3, y=0= x2-8x +15 and
x=5 then y=0= x2-8x +15.
To solve an equation in standard form (the right side of
the equation = 0) we plot the left side and find where it crosses the x axis.
Lets see how to do this with a calculator.
First we must input the equation. Going back to the x2-8x
+15=0
Use the x2 key next
We need to find the x value where our graph crosses the x
axis. To do this hit
Now the program we want is the 2nd one called
zero. A zero of a function is the value of x that will return a y of zero. Press
2 to enter this function
Our key presses to this point were:
We now have the courser at an appropriate point for the
left bound. It is definitely to the left of the zero marked by the circle but
relatively close to it. To select this left bound we press enter at the bottom
right of the calculator.
Notice the calculator is now asking for a right bound. The
right bound should be far enough to the right to be to the right of our selected
zero marked by the circle but not so far to the right as to include or be to the
right of the second zero at x=5. We use the right arrow again (not the up or
down ones) to move the curser to an appropriate right bound.
Notice I have moved my curser to the right (arrow) of the
zero I am trying to find (marked by the circle). But the curser is still to the
left of the second zero. Press enter to accept this right bound.
Your screen should look like this . Here is the key press
history so far.
Notice the two arrows at the top of the screen. They are
telling you that you have told the calculator to look for a zero, a place where
the graph crosses the x axis between these two arrows. They should point in
toward each other. I drew a box on this picture to show better where the
calculator is going to look for the zero. Notice we include the zero at x=3 in
our search but not the one at x=5. Only one zero should be included in our
search zone. Notice the calculator is now asking for a guess. It is saying help
me out. Point me in the right direction. So move the curser as close to the zero
you are searching for as you can get it.
I have used the left arrow to move the cursor back to the
zero I am looking for. Press enter to accept the appropriate guess.
The calculator has found our zero at x=3. Our Key presses
to this point have been:
Now let us find our second zero. Press:
once again and select 2
by pressing
The cursor is now to the right of our first zero and left
of the second zero, the one we are searching for. This is an appropriate left
bound so we press enter.
The cursor is now to the right of our second zero at a
point appropriate for our right bound so we press enter to accept this right
bound.
Once again I have drawn a box on the picture at left to
show our search area. Notice it includes our selected zero but not the other
zero. The calculator is asking for a guess so we move the cursor as close to the
zero we are searching for as possible. The calculator will use our guess as it's
starting point when it runs its numerical search routine. Use the left arrow to
move back for your guess.
I now have the coordinates of my second zero at x=5, y=0.
The y=0 is redundant since be the definition of zero y will equal 0. But x=5 is
the second solution to my quadratic equation. Be low is our key press history
from start to finish. Please go over this again with your calculator.
Now solve x2+x-12=0 on your calculator. The
solution x=3 is shown along with the key press history:
Find the second solution to the equation above.
Now solve x2-9x-36=0:
Notice the graph only crosses the x axis in one spot. That
is because we only see the graph through the window of the calculator screen.
There is another solution but it is outside the screen. Too see this look below.
The red rectangle represents the calculator screen above. If we could see the
entire graph we would see it crosses the x axis in two places.
In order to see the entire graph we need to change the
scale of the graph. To do this press
and you get the screen at left. We want to
raise the scale or "Zoom out" . Press 3 to zoom out.
Notice the scale did not change. There is a dot at the
origin that is flashing. The calculator is waiting for you to tell it where to
center the new view. Since the flashing dot is at the origin we press enter to
accept this center. If we wanted it to center some place else we would use the
arrow keys to move the flashing dot to the area we want for center of our new
view. Press enter if you have not done so already.
Now we can go ahead and solve for the two solutions. The
key presses for the one on the right is shown below.
Our scale is now off for the next problem. To return
to standard scale we press
For
standard zoom. Do so at this time.
Let's look at one final example we will graph and solve x2-144=0.
Enter the equation into the calculator.
We can't see the graph at all.
After zooming out once we see the screen at right and can
find the two solutions.
In some cases you may need to zoom out multiple times to
find the x intercepts. In general we can find the solution to any quadratic
equation if we can find the proper zoom to show the two intercepts. Sometimes
Sometimes there is only one intercept. This is a repeated
root and the Quadratic will be a perfect square . This one is (x-2)2=0
Sometimes there is no intercept and that means there is no
real solution . We will study this more later in the course.
