# Am going to use numerical methods to solve equations that can't be solved algebraically

Greig Boyd

C3 Coursework

Solution of equations by numerical methods

Introduction

In this coursework I am going to use numerical methods to solve equations that can’t be solved algebraically, for example if the largest power of the equation was 3 or over , we would have to use the following methods: the change in sign method, Newton-Raphson method and the method of rearranging f(x) =0 in the form x = g(x).

I am going to use an equation which cannot be factorised or solved using the formula x = -b  .

2a

Change of sign method

My equation is -6x3 +9x2 +5x-6.

My equation has four roots I am going to use change of sign to find one of them. For the change in sign method I am going to do a decimal search, i.e. I have taken the negative and positive y values that are in between zero. I did my first decimal search to two decimal places and carried on the same procedure all the way up to four decimal places.

The first thing I did was find out were the formula crosses the x-axis, which was 0.7 and 0.8. I then expanded this to one more decimal place and found the point where it crosses, to be between 0.71 and 0.72. I expand this again twice more to find the 3rd and 4th decimal places which are (0.716 & 0.717) and (0.7169 and 0.717) respectively.

Thus proves that the root lies no greater nor less than 0.00005 away from 0.717. It also shows it’s correct by the change in sign.

Failure of “change of sign”

Failure will occur if:

• The curve touches the x-axis.
• There are several roots close together
• There is a discontinuity in f(x)

I am going to use decimal search to find the roots of another equation

-2x2-3x-1.12

I cannot use decimal search for this equation as it doesn’t cross the x-axis.

There are some cases where the roots cannot always be found. Reasons for this could be if the roots are too close together

Newton-Raphson method

X5-6x +2

The Newton raphson method is a fixed point estimation method. It is important to use an estimation of the root as a starting point.

I will give an estimation (x0) for a root of f(x) =0 I will then draw a tangent to the curve y=f(x) at the point (x0, f(x)). The point where the tangent cuts the x-axis gives a closer approximation for the root, and then the process is repeated

From this it is clear that D is a better estimation than B. ...