Solve the equation: X3- 5X-5 = 0. This equation cannot be factorised nor solved algebraically therefore it must be solved through interval bisection.
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Introduction
Interval Bisection.
Ben Ward
Solve the equation: X3– 5X-5 = 0. This equation cannot be factorised nor solved algebraically therefore it must be solved through interval bisection.
To calculate the approximate position of the roots and the number of roots I used Omnigraph to sketch the graph (below shows where the graph crosses the x-axis).
This is the root I want to find and shows how the solution lies between 2.6 and 2.7

[f (X) = X3 – 5X - 5]
X0 = 2.6 f (X0) = -0.424
X1 = 2.7 f (X1) = + 1.183
As f(X0)< 0 and f(X1) > 0 it shows that the solution must lie between X0 and X1, thus creating an error bound 2.6 < X < 2.7.
In order to reduce the error bound and ultimately find the solution I set-up a table of values for X and f(X) using Excel.
Middle
1
2.62734375
11
2.6273926
0.0004319
1
2.62734375
12
2.6273682
4.837E-05
1
2.62734375
13
2.627356
-0.000143
-1
2.62736816
14
2.6273621
-4.75E-05
-1
2.62736816
15
2.6273651
4.335E-07
1
2.62736206
16
2.6273636
-2.35E-05
-1
2.62736511
After 16 literations it gives the answer of 2.627 to 3 d.p or 2.627 +/- 0.0005. Therefore it can be said that that 2.627<X<2.628.
As the sign of (X) when its equal to 2.627 is negative
i.e. 2.627 3 – 5*2.627 - 5 = -0.00573
- And the sign of (X) when it equals 2.628 is positive
i.e. 2.628 3 – 5*2.628 - 5 = -0.00998
To give an example of an equation that cannot be solved using this method is an equation that follows two main points:
- It must be an equation that just touches the x-axis but doesn’t cross the axis i.e. X4 (shown on omnigraph below).
- Must not touch the x-axis at the origin- therefore the graph of X4 needs to be ‘shifted’ either left or right i.e. (X-7)4 [shown below on omnigraph].
Conclusion
An example of this is: X3-3X2+X+1 = 0.
The diagram below shows that the gradient of the curve X3-3X2+X+1 = 0 although is not too steep to allow the Newton Raphson method to calculate the value of the root but highlights how the solution lies close to a maximum point on the graph.
It is also clear that by using excel to create a spread sheet, as used to calculate the solution for: 2X4- X3 + 3X2-2 = 0, shows how the tangent and therefore the answer moves further away and doesn’t give a correct solution.
Newton raphson | |
n | X(n) |
0 | 0.5 |
1 | 1.2 |
2 | 0.991489 |
3 | 1.000001 |
4 | 1 |
5 | 1 |
6 | 1 |
7 | 1 |
8 | 1 |
9 | 1 |
10 | 1 |
11 | 1 |
This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.
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