Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.
Graph splitting<br />- A method for localising the roots of a polynomial<br />
Example 1:<br />x3 + x2 + x + 1  =  0.           Solve for x<br />First question:     How many answers (if any) will there...
Example 1:<br />x3 + x2 + x + 1  =  0.           Solve for x<br />First question:     How many answers (if any) will there...
Example 1:<br />x3 + x2 + x + 1  =  0.           Solve for x<br />Say   x3 + 1  =  - x2 - x<br />        Set    y  =     x...
Example 1:<br />x3 + x2 + x + 1  =  0.           Solve for x<br />Say   x3 + 1  =  - x2 - x<br />        Set    y  =     x...
Example 1:<br />x3 + x2 + x + 1  =  0.           Solve for x<br />Say   x3 + 1  =  - x2 - x<br />        Set    y  =     x...
Example 2:<br />x4 - x2+ 1  =  0. <br />
Example 2:<br />x4 - x2+ 1  =  0. <br />Say   x4+ 1  =  x2<br />        Set    y  =     x4+ 1  <br />        And   y  = x2...
Example 2:<br />x4 - x2+ 1  =  0. <br />Say   x4+ 1  =  x2<br />        Set    y  =     x4+ 1  <br />        And   y  = x2...
Example 3:<br />x4 - x - 1  =  0. <br />
Example 3:<br />x4 - x - 1  =  0. <br />Say   x4 - 1  =  x<br />        Set    y  =     x4 - 1  <br />        And   y  = x...
Example 3:<br />x4 - x - 1  =  0. <br />Say   x4 - 1  =  x<br />        Set    y  =     x4 - 1  <br />        And   y  = x...
END<br />
Upcoming SlideShare
Loading in …5
×

Solving algebraic equations by 'Graph splitting'

1,902 views

Published on

A graphical technique for localising the x-intercepts of a polynomial function

Published in: Education, Technology
  • Be the first to comment

  • Be the first to like this

Solving algebraic equations by 'Graph splitting'

  1. 1. Graph splitting<br />- A method for localising the roots of a polynomial<br />
  2. 2. Example 1:<br />x3 + x2 + x + 1 = 0. Solve for x<br />First question: How many answers (if any) will there be?<br />Second question: Where should I look?<br />
  3. 3. Example 1:<br />x3 + x2 + x + 1 = 0. Solve for x<br />First question: How many answers (if any) will there be?<br />Second question: Where should I look?<br />Split the equation into two parts that can be graphed by hand easily.<br />
  4. 4. Example 1:<br />x3 + x2 + x + 1 = 0. Solve for x<br />Say x3 + 1 = - x2 - x<br /> Set y = x3 + 1 <br /> And y = - x(x+1)<br />Split the equation into two parts that can be graphed by hand easily.<br />
  5. 5. Example 1:<br />x3 + x2 + x + 1 = 0. Solve for x<br />Say x3 + 1 = - x2 - x<br /> Set y = x3 + 1 <br /> And y = - x(x+1)<br />Draw these graphs on the same axes and find the point(s) of intersection.<br />
  6. 6. Example 1:<br />x3 + x2 + x + 1 = 0. Solve for x<br />Say x3 + 1 = - x2 - x<br /> Set y = x3 + 1 <br /> And y = - x(x+1)<br />There is only one solution, <br />and its value is x = -1<br />
  7. 7. Example 2:<br />x4 - x2+ 1 = 0. <br />
  8. 8. Example 2:<br />x4 - x2+ 1 = 0. <br />Say x4+ 1 = x2<br /> Set y = x4+ 1 <br /> And y = x2<br />
  9. 9. Example 2:<br />x4 - x2+ 1 = 0. <br />Say x4+ 1 = x2<br /> Set y = x4+ 1 <br /> And y = x2<br />The lines don’t intersect. <br />There are no solutions.<br />
  10. 10. Example 3:<br />x4 - x - 1 = 0. <br />
  11. 11. Example 3:<br />x4 - x - 1 = 0. <br />Say x4 - 1 = x<br /> Set y = x4 - 1 <br /> And y = x<br />
  12. 12. Example 3:<br />x4 - x - 1 = 0. <br />Say x4 - 1 = x<br /> Set y = x4 - 1 <br /> And y = x<br />There are two solutions, <br />close to x = 1 and x = -1.<br />
  13. 13. END<br />

×