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1. Convergence of Newton-Raphson Method
Suppose 𝑥𝑛 differs from the root 𝛼by a small quantity 𝜀𝑛 such that 𝑥0 = 𝛼 + 𝜀𝑛 and 𝑥𝑛+1 =
𝛼 + 𝜀𝑛+1
Then 𝛼 + 𝜀𝑛+1 = 𝛼 + 𝜀𝑛 −
𝑓 𝛼+𝜀𝑛
𝑓′ 𝛼+𝜀𝑛
𝜀𝑛+1= 𝜀𝑛 −
𝑓 𝛼+𝜀𝑛
𝑓′ 𝛼+𝜀𝑛
= 𝜀𝑛 −
𝑓 𝛼 +𝜀𝑛𝑓′ 𝛼 +
1
2!
⋅𝜀𝑛
2𝑓′′ 𝛼 +⋯
𝑓′ 𝛼 +𝜀𝑛𝑓′′ 𝛼
=𝜀𝑛 −
𝜀𝑛𝑓′ 𝛼 +
1
2!
⋅𝜀𝑛
2𝑓′′ 𝛼 +⋯
𝑓′ 𝛼 +𝜀𝑛𝑓′′ 𝛼
=
𝜀𝑛
2
2
𝑓′′ 𝛼
𝑓′ 𝛼
This shows that the subsequent error at each step, is proportional to the square of the
previous error and as such the convergence is quadratic. Thus Newton Raphson method has
second order convergence.

2. Rule-I Descarte's Rule of sign (for real roots)
In any algebra equation 𝑓 𝑥 = 0
1. The number of positive roots cannot exceed the number of changes of signs (or variations) from positive to negative
and negative to positive in𝑓 𝑥 .
2. The number of negative roots cannot exceed the number of variations in 𝑓 −𝑥 .
If we consider the equation 2𝑥7 − 𝑥5 + 4𝑥3 − 5 = 0
Rule-II
Every equation of an odd degree has at least one real root opposite to that of (
𝑎𝑛
𝑎0
), whereas every equation of an even
degree whose last term is negative has at least two real roots, one positive and the other negative.
Rule-III Relation between roots and coefficients
If 𝛼1, 𝛼2, 𝛼3 … 𝛼n are the roots of the equation
𝑥n + 𝑎 1 𝑥 n-1 + 𝑎 2 𝑥 n-2 + . . . + 𝑎 n-1 𝑥 + 𝑎 n = 0
𝛼 1 = −
𝑎1
𝑎0
𝛼 1𝛼2 = −
𝑎2
𝑎0
𝛼 1𝛼2𝛼3 = −
𝑎3
𝑎0
Some rules to Locate Roots of 𝑓 𝑥 = 0

3. In all the previous methods we require prior information about the root, but in this method we do
not required any previous information and it is capable of giving all the roots of algebraic equation
𝑓 𝑥 = 0 at a time.
This is a direct method to find the roots of any polynomial equation with real coefficients. The
basic idea behind this method is to separate the roots of the equations by squaring the roots. This
can be done by separating even and odd powers of 𝑥 in
𝑓𝑛 𝑥 = 𝑥n + 𝑎 1 𝑥 n-1 + 𝑎 2 𝑥 n-2 + . . . + 𝑎 n-1 𝑥 + 𝑎 n = 0
Now separate odd and even degree powers of x and squaring on both sides. Thus we get,
(𝑥 n + 𝑎 2 𝑥 n-2 + 𝑎 4 𝑥 n-4 + . . . )2 = (𝑥 n + 𝑎 1 𝑥 n-1 + 𝑎 3 𝑥 n-3 + . . . )2
𝑥 2n
n - (𝑎 1
2 - 2 𝑎 2) 𝑥 2n-2 + (𝑎 2
2 - 2 𝑎 1 𝑎 3 + 2 𝑎 4) 𝑥 2n-4 +. . . +(-1)n 𝑎 n
2 = 0
Substituting y for 𝑥 2 we have
𝑦 n + 𝑏 1 𝑦 n-1 + . . . + 𝑏 n-1 𝑦 + 𝑏 n = 0
Where 𝑏 1 = 𝑎 1
2 - 2 𝑎 2
𝑏 2 = 𝑎 2
2 - 2 𝑎 1 𝑎 3 + 2 𝑎 4
𝑏 n = −1 2 𝑎 n
2
Graeffe's Root Squaring Method

4. If 𝛼1, 𝛼2, 𝛼3 … 𝛼n are the roots of original equation then 𝛼1
2
, 𝛼2
2
, 𝛼3
2… 𝛼n
2 are the roots of
the above equation.
This procedure can be repeated many times so that the final equation
𝑧 n +𝑐1 𝑧n-1 + . . . + 𝑐 n-1𝑧+ 𝑐 n= 0
has the roots
𝛾 1, 𝛾 2 , . . ., 𝛾 n such that 𝛾 i = 𝛼 i
(2^m) where 𝑖 = 1, 2, . . ., 𝑚 Type equation here. if we
repeat the process m times.
If we assume |𝛼1| > |𝛼2| > . . . |𝛼n| then | 𝛾 1| >> | 𝛾 2| >> . . .>> | 𝛾 n|
that is the roots 𝛾 i are very widely separated for large 𝑚.
Then
𝛾 1 = −𝑐1
𝛾 2 =
−𝐶2
𝐶1
𝛾3 =
−𝐶3
𝐶2
Now since 𝛾 i = 𝛼 i
(2^m) 𝛼𝑖 = 𝛾𝑖
1
2𝑚 =
𝐶𝑖
𝐶𝑖−1

5. Apply Graeffe's root squaring method to solve the equation 𝑥3−8𝑥2 + 17𝑥 − 10 = 0
Given 𝑥3
−8𝑥2
+ 17𝑥 − 10 = 0
Rearranged given equation by separating even and odd degree powers of x, we get
𝑥3
+17𝑥 = 10 + 8𝑥2
Squaring both side we get
x3
+ 17x 2
= 10 + 8𝑥2 2
𝑥6
+ 34𝑥4
+ 289𝑥2
= 64𝑥4
+ 160𝑥2
+ 100.
Now, let 𝑥2
= 𝑦 , we get
𝑦3 + 34𝑦2 + 289𝑦 = 64𝑦2 + 160𝑦 + 100.
𝑦3
+ 129𝑦 = 30𝑦2
+ 100
Again squaring both sides, we get
𝑦3
+ 129𝑦 2
= 30𝑦2
+ 100 2
𝑦6
+ 258𝑦4
+ 16641𝑦2
= 900𝑦4
+ 6000𝑦2
+ 10000
𝑦6
+ 10641𝑦2
= 642𝑦4
+ 10000
Example-1

6. Now, let 𝑦2
= 𝑧 to get
𝑧3
+ 10641𝑧 = 642𝑧2
+ 10000
Again squaring both side, we get
(𝑧3
+10641𝑧)2 = 642𝑧2
+ 10000 2
𝑧6 + 21282𝑧4 + 113230881𝑧2 = 412164𝑧4 + 12840000𝑧2 + 100000000
Putting 𝑧2
= 𝑢, we get
𝑢3 − 390882𝑢2 + 100390881𝑢 − 108 = 0
Comparing the above equation with 𝑢3 + 𝐵1𝑢2 + 𝐵2𝑢 + 𝐵3 = 0, we get
𝐵1 = −390882, 𝐵2 = 100390881, 𝐵3 = −108
.
Let the roots of equation (2)are
𝑥𝑖 = 𝑅 1/2𝑚
= −
𝐵𝑖
𝐵𝑖−1
1
2𝑚
: 𝐵0 = 1
𝑥1 = 𝑅1
1
8= −𝐵1
1
8 = 390882
1
8 = 5.004 ≈ 5
𝑥2 = 𝑅2
1
8=
−B2
B1
1
8
=
100390881
390882
1
8
= 2.000 =2
𝑥3 = 𝑅3
1
8=
−B3
B2
1
8
=
108
100390881
1
8
= 0.999 ≈ 1
Thus, roots of given equation are 5 , 2 and 1.

7. Find all roots of the equation 𝑥3-2 𝑥2-5𝑥+6=0by Graeffe’s method.
Given 𝑥3
-5𝑥 = 2 𝑥2
−6
Squaring on both sides
𝑥2 (𝑥2
-5)2 = (2 𝑥2
−6)2
Put 𝑥2 = 𝑦 to get
𝑦(𝑦-5)2 = (2𝑦 -6)2
𝑦(𝑦2 +49) = 14 𝑦2+36
Again squaring on both sides and putting 𝑦2 = 𝑧, we get
𝑧(𝑧+49)2 = (14𝑧 +36)2
Or 𝑧(𝑧2+1393) = 98𝑧2 +1296
Again squaring and putting 𝑧2 = 𝑢, we get
𝑢(𝑢+1393)2 =( 98𝑢 +1296)
Or 𝑢3
-6818 𝑢2
+ 1686433𝑢-1679616=0
If the roots of this equation are R1,R2,R3
Example-2

8. 𝑥1 = 𝑅1
1
8= −𝐵1
1
8 = 6818
1
8 = 3.0144 ≈ 3
𝑥2 = 𝑅2
1
8=
−B2
B1
1
8
=
1686433
6818
1
8
= 1.9914 =2
𝑥3 = 𝑅3
1
8=
−B3
B2
1
8
=
1679616
1686433
1
8
= 0.9994 ≈ 1
By Descartes rule of sign, there is one negative root which is found by substituting the
values in the given equation, we find
𝑓 3 = 0, 𝑓 −2 = 0 𝑓 1 = 0
Hence the roots are 3,-2,1.

9. Complex Roots
If 𝛼𝑟 and 𝛼𝑟+1form a complex pair 𝜌𝑟𝑒±𝑖𝜙,then the coefficients of 𝑥𝑛−𝑟in successive
squarings would fluctuate both in magnitude and sign by an amount (2𝜌𝑟
𝑚 cos𝑚𝜙 𝑟)
For sufficiently large 𝑚, 𝜌𝑟and 𝜙𝑟can be determined by
𝜌𝑟2 2𝑚
=
𝑐𝑟+1
𝑐𝑟−1
, 2𝜌𝑟
𝑚 cos𝑚𝜙 𝑟=
𝐶𝑟+1
𝐶𝑟−1
If the equation has only one pair of complex roots say 𝜌𝑟𝑒±𝑖𝜙 = 𝜉 + 𝑖𝜂 then we can
find all real roots .
𝜉 is given by 𝛼1 + 𝛼2 + ⋯ + 𝛼𝑟−1
+ 2𝜉 + 𝛼𝑟+2 𝛼𝑟+2 + ⋯ 𝛼𝑛 = 0
And 𝜂 is given by 𝜌𝑟
2
= 𝜉2
+ 𝜂2
𝑜𝑟 𝜂 = 𝜌𝑟
2
− 𝜉2

10. Example-1
Apply Graeffe’s method to find all the roots of the equation 𝑥4
− 3𝑥 + 1 = 0.
We have
𝑓(𝑥) = 𝑥4
− 3𝑥 + 1 = 0
+ − +
There being two changes in sign, above equation has two positive real roots and no negative
real roots.
Thus, the two remaining two roots are complex.
Rewriting (𝑖) as 𝑥4
+ 1 = 3𝑥, and squaring, we get
(𝑦2
+ 1)2
= 9𝑦
Where 𝑦 = 𝑥2
.

11. Squaring again and putting 𝑦2
= 𝑧, we obtain
𝑧 + 1 4 = 81𝑧 𝑜𝑟, 𝑧4 + 4𝑧3 + 6𝑧2 − 77𝑧 + 1 = 0
Or
𝑧4
+ 6𝑧2
+ 1 = −𝑧(4𝑧2
− 77)
Squaring once again and putting 𝑧2 = 𝑢, we get 𝑢2 + 6𝑢 + 1 2 = 𝑢 4𝑢 − 77 2
Or
𝑢4
− 4𝑢3
+ 654𝑢2
− 5917𝑢 + 1 = 0
If 𝛼1, 𝛼2, 𝛼3, 𝛼4 be the roots of the original equation, then roots of the above
equation are 𝛼1
8
, 𝛼2
8
, 𝛼3
8
, 𝛼4
8
. Thus, we have

12. 𝛼1
8
= 4 𝑖. 𝑒. 𝛼1 = 1.1892
𝛼2
8
=
654
4
= 163.5 𝑖. 𝑒. 𝛼2 = 1.891
𝛼3
8
=
5917
654
= 9.0474 𝑖. 𝑒. 𝛼3 = 1.3169
𝛼4
8
=
1
5917
= 0.00017 𝑖. 𝑒. 𝛼4 = 0.3379
From above equations, we observe that the magnitude of the coefficient
𝑐1 and 𝑐4 have become constant. This indicates that 𝛼1 and 𝛼4 are the
real roots where as 𝛼2 and 𝛼3 are a pair of complex roots.
∴ the real roots 𝛼1 = 1.1892 and 𝛼4 = 0.3379.
Now let us find the complex roots 𝜌2𝑒±𝜙2 = 𝜉 + 𝑖𝜂

13. Hence, we have
𝜌2
2 23
=
𝑐2+1
𝑐2−1
𝑜𝑟 𝜌2
16
=
5917
4
= 1479.25
Where 𝜌2 = 1.5781
Also, from original equation, 𝛼1 + 2𝜉 + 𝛼4 = 0
This gives
𝜉 = −
1
2
𝛼1 + 𝛼4 = −0.7636 𝑎𝑛𝑑 𝜂 = 𝜌2
2
− 𝜉2 = 1.381
Hence the complex roots are −0.7636 ± 1.381𝑖