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# Solving algebraic equations by 'Graph splitting'

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

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### 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 />