SOLUTION: From a worksheet. Solve 2x^2=3x+2 by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. I must show my work, and don't

Algebra ->  Graphs -> SOLUTION: From a worksheet. Solve 2x^2=3x+2 by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. I must show my work, and don't      Log On


   



Question 165511: From a worksheet.
Solve 2x^2=3x+2 by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. I must show my work, and don't understand how to do this.
Thanks

Found 2 solutions by vleith, Edwin McCravy:
Answer by vleith(2983) About Me  (Show Source):
You can put this solution on YOUR website!
Solve 2x^2=3x+2 by graphing.
2x%5E2=3x%2B2
2x%5E2+-+3x+-+2+=+0
By graphing, make a table and put in a bunch of values for x and then find the the "y" values using f%28x%29+=+2x%5E2+-+3x+-+2
graph%28400%2C400%2C+-10%2C10%2C+-10%2C10%2C+2x%5E2+-+3x+-+2%29

Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
From a worksheet.
Solve 2x%5E2=3x%2B2 by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. I must show my work, and don't understand how to do this.
Thanks

There are two ways to do this by graphing:

Method 1:

Graph y=left_side}
That is, draw the graph of y=2x%5E2


Now draw the graph of the y=right_side
That is, draw the graph of y=3x%2B2 on the
same set of axes:



From each of the two points where they cross,
draw a vertical line directly to the x-axis, 
like this:



Notice that the vertical line on the left appears
to hit the x-axis at -1%2F2 and the right
vertical line appears to hit the x-axis at 2

So It appears the roots are x=-1%2F2 and x=2.

Method 2:

2x%5E2=3x%2B2

Get 0 on the right side;

2x%5E2-3x-2=0

Temporarily replace the 0 by y:

Draw the graph of 

y=2x%5E2-3x-2



Now since we started with 2x%5E2-3x-2=0

y can  DISABLED_event_ONLY=  0, so the only valid points on
that graph are the two points which are ON
the x-axis.  That's because every point on the 
x-asis has 0 for its y-coordinate.  Also the
points on the x-axis are the only ones which
have their y-values 0.

So the two roots appear to be x=-1%2F2 and y=2.

I'm not sure which method your teacher wants.  Either
is correct, but be sure to ask your teacher.

Edwin