ROOTS OF ALGEBRAIC AND TRANSCENDENTAL EQUATIONS
Generally an equation is solved by factorization.
But in many cases, the method of factorization fails.
In such cases, numerical methods are used.
The numerical methods to solve the equation f(x) = 0 are:
1) BISECTION METHOD
2) REGULA FALSI METHOD
3) NEWTON RAPHSON METHOD
4) SECANT METHOD
1) BISECTION METHOD
Bisection Method is also called the interval halving method, the binary search method, or the dichotomy method.
The Rate of Convergence of Bisection Method is Linear.
For any continuous function f(x),
Find two points say a and b such that f(a) < 0 and f(b) > 0
Find midpoint of a and b say x_{n}
x_{n }= (a + b)/2
Question :
Obtain the root of the equation x – cosx = 0 by bisection method correct up to two decimal places.
Solution :
f(x) = x – cosx
f(0) = 1 and f(1) = 0.4597
Since f(0) < 0 and f(1) > 0 , the root lies between 0 and 1.
No. 
A 
B 
x_{n} = (a + b)/2 
Sign of f(x_{n}) 
1 
0 
1 
0.5 
ve 
2 
0.5 
1 
0.75 
+ve 
3 
0.5 
0.75 
0.625 
ve 
4 
0.625 
0.75 
0.6875 
ve 
5 
0.6875 
0.75 
0.71875 
ve 
6 
0.71875 
0.75 
0.7344 
ve 
7 
0.7344 
0.75 
0.7422 
+ve 
8 
0.7344 
0.7422 
0.7383 
ve 
9 
0.7383 
0.7422 
0.74025 
+ve 
10 
0.7383 
0.74025 
0.7393 
+ve 
11 
0.7383 
0.7393 
0.7388 

Since x_{10} and x_{11} are same up to two decimal places, the root is 0.73
2) REGULA FALSI METHOD
Regula Falsi Method is also known as False Position Method.
x_{n }= [af(b) – bf(a)]/[f(b) – f(a)]
Question:
Find the real root of x log_{10} x – 1.2 = 0, correct up to three decimal places.
Solution:
f(x) = x log_{10} x – 1.2
f(2) = 0.5979 and f(3) = 0.2314
Since f(2) < 0 and f(3) > 0, the root lies between 2 and 3.
No. 
a 
b 
f(a) 
f(b) 
x_{n }=[af(b) – bf(a)]/[f(b) – f(a)] 
f(x_{n}) 
1 
2 
3 
0.5979 
0.2314 
2.721 
0.0171 
2 
2.721 
3 
0.0171 
0.2314 
2.7402 
0.0004 
3 
2.7402 
3 
0.0004 
0.2314 
2.7406 

The root is 2.740
3) NEWTONRAPHSON METHOD
The Newton Raphson Method has a quadratic convergence and the convergence is of the order of 2.
Newton raphson formula or Newton’s iteration formula is :
x_{n+1} = x_{n} – [f(x_{n})/f’(x_{n})]
Question:
Find the real positive root of the equation x – cosx = 0 using Newton Raphson Method correct up to three decimal places.
Solution:
f(x) = x – cosx
f(0) = 1 and f(1) = 0.4597
Since f(0) < 0 and f(1) > 0 , the root lies between 0 and 1.
Let x_{0} = 1
f’(x) = 1 + sinx
f(x_{0}) = f(1) = 0.4597
f^{’}(x_{0}) = f’(1) = 1.8415
x_{1} = x_{0} – [f(x_{0})/f’(x_{0})]
= 1 – (0.4597/1.8415)
= 0.7504
f(x_{1}) = f(0.7504) = 0.019
f^{’}(x_{1}) = f’(0.7504) = 1.6819
x_{2} = x_{1} – [f(x_{1})/f’(x_{1})]
= 0.7504 – (0.019/1.6819)
= 0.7391
f(x_{2}) = f(0.7391) = 0.00002
f^{’}(x_{2}) = f’(0.7391) = 1.6736
x_{3} = x_{2} – [f(x_{2})/f’(x_{2})]
= 0.7391 – (0.00002/1.6736)
= 0.7391
Since x_{2} and x_{3} are same up to three decimal places, the root is 0.739
4) SECANT METHOD
x_{n+1 }= x_{n} – {[(x_{n } – x_{n1}) * f(x_{n})]/ [f(x_{n}) – f(x_{n1})]}
The rate of convergence of secant method is 1.618.
The rate of convergence of secant method is lesser than NewtonRaphson Method and so Secant Method is more efficient than the Newton Raphson Method.
Question:
Find the root of xlog_{10} x – 1.9 = 0, correct up to three decimal places with x_{0} = 3 and x_{1 }= 4.
Solution:
f(x) = xlog_{10} x – 1.9
x_{0 }= 3, x_{1} = 4
f(x_{0}) = f(3) = 0.4686, f(x_{1}) = f(4) = 0.5082
x_{2 }= x_{1} – {[(x_{1 } – x_{0}) * f(x_{1})]/ [f(x_{1}) – f(x_{0})]}
= 4 – {[4 – 3) * (0.5082)]/[(0.5082 + 0.4686)]}
= 3.4797
f(x_{2}) = f(3.4797) = 0.0156
x_{3 }= x_{2} – {[(x_{2} – x_{1}) * f(x_{2})]/ [f(x_{2}) – f(x_{1})]}
= 3.4979 – {[3.4979 – 4) * (0.0156)]/[(0.0156 – 0.5082)]}
= 3.4952
f(x_{3}) = f(3.4952) = 0.0005
x_{4 }= x_{3} – {[(x_{3 } – x_{2}) * f(x_{3})]/ [f(x_{3}) – f(x_{2})]}
= 3.4952 – {[3.4952 – 3.4797) * (0.0005)]/[(0.0005 + 0.01556)]}
= 3.4957
Since x_{3} and x_{4} are same up to three decimal places, the root is 3.495