CHAPTER TWO

Solution of Equations
3.1- Introduction:
Finding the roots of equations is one of the oldest problems in mathematics,
since it is required in a great variety of applications.
Consider the simple quadratic equation:
ax2 +bx + c = 0
−𝑏± 𝑏 2 −4𝑎𝑐
we say that x = ….. (3.1)
2𝑎

are the roots of this equation, because for these values of (x), the quadratic
equation is satisfied.
Functions can be divided into two types:
1. Polynomials: which can be written in the form of:
F(x) = a0 + a1x + a2x2 + a3x3 + a4x4+ ……. +anxn
Example : f(x) = 1 - 2.37 x + 7.5 x2
And f(x) = 5x2 - x3 + 7 x6
2. Transcendental Functions: that is non-algebraic, which include
trigonometric , exponential, logarithmic . . etc.
Example : f(x) = e-x - x
F(x) = Sin x
F(x) = ln x2 – 1
3.2- Descart's rule of signs:

The number of positive real roots of fn(x) = 0 cannot exceed the number of (+ve )
sign changes in fn(x). And the number of (-ve) real roots of fn(-x).

Example 3.1: Find the number of positive and negative real roots in the equation:
f5(x) = 8x5 + 12x4 - 10x3 + 17x2 - 18x +5 = 0

Solution: Let x=1 then f5(1) = +8 +12 -10 +17 -18 +5

∴number of (+ve) real roots are (four).
Let x = -1 then f5(x) = -8 + 12 + 10 + 17 + 18 +5
∴number of (-ve) real roots is (one).
>> p1=[8 12 -10 17 -18 5]

p1 =
8 12 -10 17 -18 5
>> roots_p1=roots(p1)
roots_p1 =
-2.5000
0.0000 + 1.0000i
0.0000 - 1.0000i
0.5000 + 0.0000i
0.5000 - 0.0000i
Example 3.2: Use Descart's rule of signs to the number of real roots of: f(x) =
x5 + x4 + 4x3 + 3x2 + x + 1
Solution: Let's find first F(1) = +x5 + x4 + 4x3 + 3x2 + x + 1
There are no sign changes, so there are no (+ve) real roots.
Now let's find f(-1) = - x5 + x4 - 4x3 + 3x2 - x + 1
There are (5) sign changes, so there are five (-ve) roots.

Example 3.3: f(x) = 2x4 – x3 + 4x2 – 5x + 3

Example 3.4: f(x) = x5 – 3x3 + 8x – 10
Sol:
>> f1=[1 0 -3 0 8 -10]
f1 =

1 0 -3 0 8 -10

>> roots_f1=roots(f1)

roots_f1 =

-1.6792 + 0.9601i
-1.6792 - 0.9601i
1.5383 + 0.0000i
0.9100 + 0.9536i
0.9100 - 0.9536i
Example 3.5: f(x) = 6x3 + 7x2 – 3x + 1

>> f1=[6 7 -3 1];

>> roots_f1=roots(f1)
roots_f1 =
-1.5566 + 0.0000i
0.1950 + 0.2628i
0.1950 - 0.2628i
3.3- Methods of Solutions: There are two groups of methods to solve the
functions: - Bracketing methods
- Open methods

3.3.1- Bracketing methods: It is named because all the methods used within a
certain interval (or range) of the function. It includes:
- Graphical method
- Bisection method
- False position method

3.3.1.1- Graphical method:

It is a simple method to find the roots, by making a table for assumed values of
(x)for a certain range (min. value is called (xlower or (xl) and max. value that is call
xupper (xu)), they are substituted in the function to find the value of (f(x) or y), then
making a plot of the function and observe where it crosses the x-axis (i.e. root).
Graphical techniques are of limited practical value, because they are not precise.

Fig. 3.1
 In fig 3.1a,c , indicate that f(xl) and f(xu)have the same sign, either there will be
no roots, or there will be even number of roots within the interval.
In fig 3.1 b, d, indicates that the function have different signs at the end points,
so there will be an odd number of roots in the interval.
Generally if f(xu) * f(xl) < 0 , means there is at least one root between them.

3.3.1.2- Bisection method: It is one type of incremental search method which

the interval is always divided in half. The location of the root is then determined
as laying at the midpoint of the subinterval within which the sign change occurs.
The following are the steps for actual computation:
Step #1: Choose the lower (xl=a) and upper (xu=b) guesses for the roots so that the
function change sign.
Step #2: An estimate of the root (xm=c) is determined by:
𝑥𝑙 + 𝑥𝑢
𝑥𝑚 = … … … … . (3.2)
2
Step #3: Make the following evaluation to determine which subinterval the root is:
1. F(xl)*f(xm)<0 the root lies in the lower subinterval
2. F(xl)*f(xm)>0 the root lies in the upper subinterval
3. F(xl)*f(xm)=0 the root equals (xm)[terminate computation]
Example3.6 : Use the method of bisection to find the root of the equation, f(x) =
x 4 +2x 3 -x-1 = 0 lying in the interval [0, 1] at the end of sixth iteration. How
many iterations are required if the permissible error is Єs = 0.0005 ?

Sol. :Assume xl = a , xu = b and xm = c

The given interval is [a, b] = [0, 1]
Iteration No. 1 a= 0, b= 1
𝑎+𝑏 0+1
∴ c = = =0.5
2 2
f(a) = f(0) = -1,
f(b) = f(1) = 1,
f(c) = f(0.5) = -1.1875
Since f(b) * f(c) < 0, root lies between 'b' and 'c'. Hence 'a' will be replaced by 'c'.
Therefore new interval is [a, b] = [0.5, 1]
0.5+1
Iteration No. 2 a= 0.5, b =1 .'. c = = 0.75
2
f(a) = f(0.5)=-1.1875,
f(b) = f(l) = 1,
f(c) = f(0.75) = - 0.5898
Since f(b) * f(c)< 0, root lies between 'b' and 'c'. Hence 'a' will be replaced by
'c', and new interval will be [a, b] = [0.75, 1]
0.75+1
Iteration No. 3 a = 0.75, b = 1 .-. c = = 0.875
2
f(a) = f(0.75) = -0.5898,
f(b) = f(1) = 1,
f(c) = f(0.875) = 0.0510254
f(a) * f(c) < 0, root lies between 'a' and 'c'
'b' will be replaced by ' c' and New interval [a, b] = [0.75, 0.875]
0.75+0.875
Iteration No. 4 a= 0.75, b = 0.875 .-. c = = 0.8125
2
f(a) = f(0.75) = -0.5898
f(b) = f(0.875) = 0.0510254,
f(c) = f(0.8125) = - 0.3039398
f(b) * f(c) < 0, root lies between 'b' and 'c'
'a' will be replaced by 'c' and new interval [a, b] = [0.8125, 0.875]
0.8125+0.875
Iteration No. 5 a = 0.8125, b = 0.875 , c = = 0.84375
2
f(a) = f(0.8125) = - 0.3039398,
f(b) = f(0.875) = 0.0510254,
f(c) = f(0.84375) = - 0.1355733
f(b) * f(c) < 0, root lies between 'b' and 'c'
'a' will be replaced by 'c' and new interval [a, b] = [0.84375, 0.875]
0.84375 +0.873
Iteration No. 6 a = 0.84375, b = 0.875, c = = 0.859375
2
f(a) = f(0.84375) = - 0.1355733,
f(b) = f(O.875) = 0.0510254,
f(c) = f(0.859375) = - 0.0446147
f(b) * f(c) < 0, root lies between 'b' and 'c'
'a' will be replaced by 'c' and new interval [a, b] = [0.859375, 0.875]
0.859375 +0.875
Hence root will be = = 0.8671875
2
Thus root = 0.8671875 at the end of 6th iteration.

To determine number of iterations for permissible error Єs = 0.0005, for the

given interval [a, b] = [0, 1]
Permissible error Єs = 0.0005
`Number of iterations are given from the equation:
log 𝑏−𝑎 −log ⁡
(∈𝑠 ) log 1−0 –log ⁡
(0.0005) 0−(−3.30103)
n≥ ≥ ≥ ≥ 10.96
log 2 log 2 0.30103
Hence (n = 11) iterations are required to get the error less than permissible error.

3.3.1.3- False position method or Regula Falsi method: It is an improved

version of the bisection method. An alternative way from halving the distance is to
join the points by a straight line. The intersection of this line with the x-axis
represents an improved estimate of the root.
From the figure, the intersection of the straight line with the x-axis can be
estimated as in the formula according to the two symmetrical triangles:

𝑦 𝑓 𝑥𝑙 𝑓 𝑥𝑢
= =
𝑥 𝑥 𝑚 −𝑥 𝑙 𝑥 𝑚 −𝑥 𝑢

𝑓 𝑥 𝑢 ∗(𝑥 𝑙 −𝑥 𝑢 )
∴xm = 𝑥𝑢 − …………………….(3.3)
𝑓 𝑥 𝑙 − 𝑓(𝑥 𝑢 )
This is the false position formula

Example 3.7:
Find the root of f(x) = e x - 4x = 0 using False position method, correct
to three decimal places.
Solution: f(x) = e x - 4x = 0, f(0)=1, f(1)=-1.281718
Hence, since f(0) * f(1) < 0 , ∴ 𝑡ℎ𝑒 𝑟𝑜𝑜𝑡 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 (0 𝑎𝑛𝑑 1)
Let's take xl = x0 = 0, and xu = x1 = 1
Using the following relation we can find the next approximation of the root;
𝑥1− 𝑥0
𝑥2 = 𝑥1 − 𝑓(𝑥1 )
𝑓 𝑥 1 −𝑓(𝑥 0 )

Iteration No.1
1−0
𝑥2 =1− ∗ −1.281718 = 0.438266
−1.281718 − 1
f(𝑥2 ) = e0.438266 - 4*(0.438266) = - 0.203047

Since f(x0) * f(x2) < 0, then the root lies between [0, 0.438266].
Hence we take initial approximation for second iteration as,
𝑥1 = 0 , 𝑥2 = 0.438266
Iteration No. 2:
With initial approximations of 𝑥1 = 0 and 𝑥2 = 0.438266 from the
previous iteration, we find next approximation x3 to the root as:

𝑥2 − 𝑥1
𝑥3 = 𝑥2 − 𝑓(𝑥2 )
𝑓 𝑥2 − 𝑓(𝑥1 )
0.438266 − 0
𝑥3 = 0.438266 − ∗ −0.203047 = 0.364297
−0.203047 − 1
f(𝑥3 ) = e0.364297 - 4*(0.364297) = - 0.017686

Since f( 𝑥1 ) * f( 𝑥3 ) < 0, root lies in the interval [0, 0.364297]

Iteration No.3:
𝑥3− 𝑥2
𝑥4 = 𝑥3 − 𝑓 𝑥3
𝑓 𝑥3 − 𝑓(𝑥 2 )

0.364297 − 0
𝑥4 = 0.364297 − −0.017686 = 0.357966
−0.017686 − 1
∴ f( 𝑥4 ) = -0.001447
Since f( 𝑥2 ) * f( 𝑥4 ) < 0, root lies in the interval [0, 0.357966]
Iteration No. 4: 𝑥5 = 0.357449
Since three decimal digits repeat in successive approximation, the
approximation to the root is correct up to 3 decimal places.
∴ 𝐴𝑛𝑠𝑤𝑒𝑟 = 0.357449
A comparison between the two methods (Bisection and False- position). It
is noted that the error for false position decreases much faster than for
bisection because of the more efficient scheme for root location in the false -
position method.

3.4- Home work:

Determine the real roots of the functions:

1. F(x) = -0.9 x 2 + 1.7x + 2.5 [take x l =2.8, x u =3.0]
2 3
2. F(x) = -2 + 6.2x – 4x + 0.7x [take x l =0.4, x u =0.6]
1. Graphically
2. Using quadratic formula (for (1)).
3. Bisection method.
4. False-position.
Compute the estimated error
5. F(x) = 2x – log 10 (x) – 7 = 0. Ans.:0.567203
3
6. F(x) = x – 2x – 5 = 0. Ans.:2.09451
Adding and subtracting X u f(X u )- X u f(X u )
3.5- Open methods:
Previous methods called bracketing methods, because it takes only a
short bounded part of the graph and check for roots. There are other
methods that is called "Open Methods" that require a single starting value or
two values that are not necessary bracket a root. We shall study three of
them:
1. - Simple One-Point Iteration
2. - Newton-Raphson method
3. - Secant method

3.5.1-Simple one-point iteration(Fixed- point iteration): By

modifying or rearranging the function so that variable (x) is on the left -hand
side of the equation
X = g(x)
This will provide a formula to predict a value of (x) as a function of
(x).Thus, given an initial guess at the root (xi) in above formula can be
used to compute a new estimate (xi+1), as expressed by the iterative
formula

X i+1 = g(x i) . . . (3.4)

The percentage relative error for this equation can be determined using the
error estimator

𝑥𝑖+1 − 𝑥𝑖
Єr% =
𝑥𝑖+1
∗ 100%

There are two ways to modify function equation to be in the form of (3.4):
1. By algebraic manipulation as: f(x) = x 2 -2x+3=0
𝑥 2 +3
∴𝑥=
2
2. By adding (x) to both sides as: f(x)=Sin(x)=0 then
x=Sin(x) + x.

Note :- |g’(x)|<1
Example 3.8:Use simple one-point iteration to locate the root

F(x) = e -x –x.
iteration 𝑿𝒊 Єa%
Solution: The function can be expressed in
0 0
the form of (3.4)as:
1 1.00 100%
𝑥𝑖+1 = 𝑒−𝑥 𝑖 2 0.367879 171
Starting with an initial guess of ( 𝑥0 = 0) 3 0.692201 46.9
to get 𝑥1 = 1, as the first iteration, then 4 0.500473 38.3
substitute again to get 𝑥2 and so on. 5 0.606244 17.4
The result is as in the table. Notice that 6 0.545396 11.2
with each iteration it becomes closer to the 7 0.579612 5.9
real root (0.56714329). 8 0.560115 3.48

Home Work:

Recalculate the previous error to find (Є t %) since you have the true
value.

3.5.2- Newton-Raphson method (NRM):

In the figure it can be seen that an initial guess at point (x i), a tangent
can be extended from the point[x i , f(x i )]. This point where this tangent
crosses the x-axis usually improved estimate of the root.
The basic formula:
𝑓 𝑥𝑖 − 0
𝑓 ′ 𝑥𝑖 =
( 𝑥 𝑖 − 𝑥 𝑖+1 )

Which can be rearranged and rewritten as:

𝑓(𝑥 𝑖 )
𝑥𝑖+1 = 𝑥𝑖 − …………………………. (3.5)
𝑓′ (𝑥 𝑖 )
This is called Newton-Raphson formula.
Example 3.9: Use the (NRM) to estimate the root of
( f(x) = e -x – x ) employing an initial guess of 𝑥0 = 0.

Solution: The first derivative of the function can be evaluated as :

𝑓 ′ 𝑥 = −𝑒 −𝑥 − 1
Which can be substituted along with the original function in the Newton -
Raphson formula to give:

𝑒 −𝑥𝑖 − 𝑥𝑖
𝑥𝑖+1 = 𝑥𝑖 −
– 𝑒 𝑥𝑖 − 1
Starting with an initial guess of 𝑥0 =0, the result will be as in the table. It is
noted that the approach rapidly converge on the true root. Notice that the
relative error at each iteration decreases much faster than it does in simple
one-point iterative method.

iteration 𝑿𝒊 Єt%
0 0 100%
1 0.5 11.8
2 0.566311 0.147
3 0.567143 0.22*10−4
4 0.5671432 <10−8

Note:
Two points to be followed using(NRM):
1. 𝑓 ′ (𝑥𝑖 ) ≠ 0. If it is "0", change value of 𝑥0 .
2. For better convergence, select 𝑥0 such that:
𝑓 𝑥0 ∗ 𝑓 ′′ 𝑥0 > 0
Example 3.10:

Find the real root of the equation:𝑥 3 + 2𝑥 − 5 = 0, by applying (NRM) for

five iterations. xsin(x)+cos(x)=0
Solution:
We have𝑓 𝑥 = 𝑥 3 + 2𝑥 − 5 = 0 ,
𝑓 ′ 𝑥 = 3𝑥 2 + 2
𝑓 ′′ 𝑥 = 6𝑥
f(0) =0 + 0 -5 = -5,
f(1)=1 +2 -5 = -2,
f(2)= 2 3 +2*2 -5 =7
Since F(1) * f(2) < 0, the root lies between 1 & 2.
For better convergence select initial value of 𝑥0 such that:
𝑓 𝑥0 ∗ 𝑓 ′′ 𝑥0 > 0
Here, 𝑓 2 ∗ 𝑓 ′′ 2 = 7 ∗ 12 > 0
So we select X 0 = 2.
𝑓(𝑥 𝑖 )
Using NRM formula: 𝑥 𝑖+1 = 𝑖 − 𝑓′ (𝑥 ) 𝑥
𝑖
𝑓 𝑥0 7
Iteration No.1: 𝑥1 = 𝑥0 − ′ =2− = 1.5
𝑓 𝑥0 14

𝑓 𝑥1 1.375
Iteration No.2: :𝑥2 = 𝑥1 − ′ = 1.5 − = 1.342857
𝑓 𝑥1 8.75
𝑓 𝑥2 0.107242
Iteration No.3: 𝑥3 = 𝑥2 − ′ = 1.342857 − 7.409796
𝑓 𝑥2
= 1.328384

𝑓 𝑥3 0.000841
Iteration No.4: 𝑥4 = 𝑥3 − ′ = 1.328384 − 7.293813
𝑓 𝑥3
= 1.328269
𝑓 𝑥4 0.000007
Iteration No.5: 𝑥5 = 𝑥4 − ′ = 1.328269 − 7.292895
𝑓 𝑥4
= 1.328269
th
Hence the root at the end of 5 iteration is correct to 6 decimal places.
Home work 1
Find the real root of the equation: xsin(x)+cos(x)=0, by applying (NRM) for
five iterations.

Example
3.5.3- The Secant method:
Some functions are difficult to evaluate its derivatives. For these cases
the derivatives can be approximated by a finite divided difference:

𝑓( 𝑥𝑖−1 ) − 𝑓(𝑥𝑖 )
𝑓′ 𝑥 =
( 𝑥𝑖−1 − 𝑥𝑖 )
This approximation can be substituted in equation (3.5) to yield the
following iterative equation:

[ 𝑥 𝑖−1 − 𝑥 𝑖 ]
𝑥𝑖+1 = 𝑥𝑖 − [𝑓( 𝑥 𝑖−1 )− 𝑓(𝑥 𝑖 )]
∗ 𝑓(𝑥𝑖 ) . . . . . .(3.6)

This is called the secant formula, which needs two arbitrary initial estimates
of ( x ).

𝑥𝑖+1
Example 3.10: Use the secant method to estimate the root of:
𝑓 𝑥 = 𝑒 −𝑥 − 𝑥 with initial estimates of [x -1 =0 and x 0 =1].
Solution:
First iteration: x -1 = 0, f(x -1 ) =1
x 0 = 1.0, f(x 0 ) = -0.63212

[ 𝑥𝑖−1 − 𝑥𝑖 ]
𝑥𝑖+1 = 𝑥𝑖 − ∗ 𝑓(𝑥𝑖 )
[𝑓( 𝑥𝑖−1 ) − 𝑓(𝑥𝑖 )]
0 − 1
∴ 𝑥1 = 1 − −0.63212 = 0.6127
1 – −0.63212

Second iteration:
x 0 =1, f(x 0 )= -0.63212
𝑥1 = 0.6127, f(𝑥1 ) = -0.07081
1 − 0.6127
∴ 𝑥2 = 0.6127 − ∗ −0.07081 = 0.56384
−0.63212 − (−0.07081)
Third iteration:
𝑥1 = 0.6127, f(𝑥1 ) = -0.07081
𝑥2 = 0.56384, f(𝑥2 ) = 0.00518

0.6127 − 0.56384
∴ 𝑥3 = 0.56384 − ∗ 0.00518 = 0.56717
−0.07081 − 0.00518

If the exact root is (0.56714329) so the percentage absolute true error (Є t )

will be:

𝑥𝑖+1 − 𝑥𝑖
Єt% =
𝑥𝑖+1
∗ 100%

1 stiteration
0.56714329 −0.6127
Єt% = ∗ 100% = 8.03%
0.56714329

By the same way the

2 nd iteration Є t % =0.58% &
 3 rd iteration Є t % = 0.0048%

Example 1
x
Determine the real root of f ( x)  e  x :
a) Graphically.
b) Using the bisection method (three iterations).
c) Using the secant method (three iterations).
d) Using the false position method (three iterations).
e) Using the Newton-Raphson method (three iterations).
Solution. a) The graphical approach for determining the roots of an equation.
x f(x)

0 1

0.1 0.804837 Root of X

1.2
0.2 0.618731
1

0.3 0.440818 0.8

0.6
0.4 0.27032 0.4
0.2
0.5 0.106531 0
-0.2 0 0.2 0.4 0.6 0.8 1 1.2
0.6 -0.05119
-0.4
-0.6
0.7 -0.20341
-0.8
0.8 -0.35067

0.9 -0.49343

1 -0.63212

The root is x  0.55; f ( x)  0.02695  0 .

b) Bisection method.
Using bisection, the results can be summarized as
Iteration, i ai bi xi f(ai) f(bi) f(xi)
0 0 1 0.5 1 -0.63212 0.106531
1 0.5 1 0.75 0.106531 -0.63212 -0.277633
2 0.5 0.75 0.625 0.106531 -0.277633 -0.089738
3 0.5 0.625 0.5625 0.106531 -0.089738 0.007283
Thus, after three iterations the root is x  0.5625, f(x) = 0.007283  0,
0.5625  0.625
a   100%  11.1% .
0.5625
c) Secant method.
Use the secant method to find the root with initial estimates of x1  0 and
x0  1.0 .
First iteration:
x1  0; f ( x1 )  1.0; x0  1; f ( x0 )  0.63212.
 0.63212  (0  1)
x1  1   0.61270; f ( x1 )  0.070814.
1  (0.63212)
Second iteration:
x0  1; f ( x0 )  0.63212; x1  0.61270; f ( x1)  0.070814.
Note that both estimates are now on the same side of the root.
 0.070814  (1  0.61270)
x2  0.61270   0.56384;
 0.63212  (0.070814)
0.56384  0.61270
f ( x2 )  0.0051798;  a   100%  8.7%.
0.56384
Third iteration:
x1  0.6127; f ( x1)  0.070814; x2  0.56384; f ( x2 )  0.00518.
0.00518  (0.6127  0.56384)
x3  0.56384   0.567696;
 0.070814  (0.00518)
0.567696  0.56384
f ( x3 )  8.66 10 4 ;  a   100%  0.7%.
0.567696
a) False position method.
Use the false position method with guesses of a0 = 0 and b0 = 1.
First iteration:
a0  0; f (a0 )  1; b0  1; f (b0 )  0.63212.
 0.63212 1
p0  1   0.61270; f ( p0 )  0.070814.
 0.63212  1
Second iteration:
Therefore, the root lies in the first subinterval, and p1 becomes:
a1  0; f (a1 )  1; b1  0.61270; f (b1 )  0.070814.
 0.070814  (0.6127)
p1  0.6127   0.57218; f ( p1 )  0.007888.
1  (0.070814)
0.57218  0.6127
a   100%  7.1% .
0.57218
Third iteration: f (a1 )  f ( p1 )  0 .
Therefore, the root lies in the first subinterval:
a2  0; f (a2 )  1; b2  0.57218; f (b2 )  0.007888.
 0.007888  (0.57218)
p2  0.57218   0.567703;
1  (0.007888)
0.567703  0.57218
f ( p2 )  8.771 10 4 ;  a  100%  0.8%.
0.567703
b) Newton-Raphson method.
x
The first derivative of the function f ( x)  e  x can be evaluated as
f ( x)  e x  1 which can be substituted along with the original function into
equation (see table 4.2) to give:

e  xi  xi
xi 1  xi  .
 e  xi  1
Starting with an initial guess of x0 = 1, this iterative equation can be applied to
compute
i xi f(xi) a, %
0 1 -0.63212
1 0.537883 0.0461 85.9
2 0.566987 0.000245 5.1
 3 0.567143 4.54110-8 2.810-2

Thus, the method rapidly converges on the true root.

Notice that the percent relative error decreases at each iteration much faster than
it does in another methods.

Home work:

1. Determine the smallest positive real root of:

𝑓 𝑥 = 𝑥 3 − 4.8𝑥 2 + 7.56𝑥 − 3.92
a. Graphically
b. Using most efficient method, employ initial guesses of:
xl=𝑥𝑖−1 =0.5 and x u = 𝑥𝑖 =1.5 and perform the computation to
within Є s = 15%.
2. Determine the roots of:
𝑓 𝑥 = 𝑥 3 − 7𝑥 2 − 3.75 + 12.5
a. Graphically
b. Using the most efficient method to within Єs = 0.1%.

2
3. Determine the real roots of f ( x)  ln x  x  7 x  8 :
(a) graphically,
(b) using the Newton-Raphson method, and
(c) using the secant method. Compare and discuss the rate of
convergence.
2
4. Locate the first positive root of f ( x)  sin( x)  x  9 x  14 . Use
four iterations of
(a) the bisection method, and
(b) the false position method.
Discuss and also perform an error check of your final answer.
5.The location 𝑥 of the centroid of a circular sector is given by :
2𝑟𝑠𝑖𝑛𝜃
𝑥=
3𝜃 𝑟
Determine the angle 𝜃 for which 𝑥 = .
2

First, derive the equation that must be solved and then determine the
root using the following methods: