Question

Example 1. Use False-Position method to find a real root of f(x) = x3 - 2x - 5 = 0 correct to
three decimal places.
(H.P.U. 2008, 09)
Sol Given equation is​

Answer 1

Step-by-step explanation:

Given . Use False-Position method to find a real root of f(x) = x3 - 2x - 5 = 0 correct to three decimal places.

• So we have
•                        f(x) = x^3 – 2x – 5 = 0
•   First we need to find the root so we have
•                        f(1) = 1^3 – 2(1) – 5  
•                              = 1 – 2 – 5
•                              = - 6
•                         f(2) = 2^3 – 2(2) – 5
•                                = 8 – 4 – 5
•                                = - 1
•                        f(3) = 3^3 – 2(3) – 5
•                                = 27 – 6 – 5
•                                 = 27 – 11
•                                 = 16
• So the root lies between 2 and 3 (one negative and other positive)
• Now using false position method
•                Now x1 = 2 and x2 = 3
•         f(x1) = f(2) = - 1
•        f(x2) = f(3) = 16
•  So x3 = x1 –  x2 – x1 / f(x2) – f(x1) x f(x1)
•       x3 = 2 -  (3 – 2) / 16 – (- 1) x -1
•          = 2 -  1 / 17 (-1)
•           = 2 + 1/17
•            = 35 / 17
•            = 2.0588
• Now f(2.0588) = (2.0588)^3 – 2(2.0588) – 5
•                          = - 0.39105
• So the root lies between 2.0588 and 3
•    Now x4 = 2.0588 – (3 – 2.0588 / 16 – (- 0.39105)
•                   = 2.08125
• Next will be x5 = 2.0862
• So repeating the process we get x9 = 2.0943 which is correct to three decimal places.

Reference link will be

https://brainly.in/question/33160190

Answer 2

Step-by-step explanation:

Given:-

Use False-Position method to find the real roots of f(x)=

${x}^{3} - 2x - 5$

First, we need to find the root so we have,

f(1)=

${1}^{3} - 2(1) - 5$

= -6

f(2)=

${2}^{3} - 2 (2) - 5$

=-1

f(3)=

${3}^{3} - 2(3) - 5$

=16

So the roots lies between 2 and 3 (one is positive and other is negative).

Take a = 2, b = 3.

  x1 = ( )– ( )

( )– ( )

af b bf a

f b fa

= 2(16) – 3(–1)

16 – (–1)

=

35

17

= 2.058823529.

f (x1) = f (2.058823529) = – 0.390799917 < 0.

Therefore the root lies between 0.058823529 and 3. Again, using the formula, we get the

second approximation as,

x2 = 2.058823529(16) – 3(–0.390799917)

16 – (–0.390799917)

= 2.08126366

Proceeding like this, we get the next approximation as,

x3 = 2.089639211,

x4 = 2.092739575,

x5 = 2.09388371,

x6 = 2.094305452,

x7 = 2.094460846