Question

1. Using the Bisection method , find the negative root of the equation X2-4X +9 =0​

Answer 1

Step-by-step explanation:

see your answer write down

Answer 2

Answer:

The required root lies between 5 and 6

Step-by-step explanation:

Given: The negative root of the equation $x^{2} - 4x + 9$

To find: The required negative root.

Solution:

Bisection method:

• A polynomial equation's roots are found using the bisection method. It splits the interval in which the equation's root is located and separates it.
• The continuous functions intermediate theorem serves as the foundation for this approach.

Let f(x) = $x^{2} - 4x + 9$

Sub x = -x

f(x) = $-x^{2} + 4x + 9$

The negative root of f(x) = 0 is the positive root of f(-x) = 0.

Then

g(x) = f(-x) = 0

g(x) = $x^{2} - 4x + 9$

Sub x = 2

⇒ g(x) = $x^{2} - 4x + 9$

⇒ g(2) = $(2)^{2} - 4 (2) -9$

          = 4 - 8 - 9

          = - 13

g(2) = - 13 which is a negative value

Sub x = 3

⇒ g(x) = $x^{2} - 4x + 9$

⇒ g(3) = $(3)^{2} - 4 (3) -9$

           = 9 - 12 - 9

           = - 12

g(3) = -12 which is a negative value

Sub x = 4

⇒ g(x) = $x^{2} - 4x + 9$

⇒ g(4) = $(4)^{2} - 4 (4) -9$

           = 16 - 16 - 9

           = -9

Sub x = 5

⇒ g(x) = $x^{2} - 4x + 9$

⇒ g(5) = $(5)^{2} - 4 (5) -9$

           = 25 - 20 - 9

           = -4

g(5) = -4 which is a negative value

Sub x = 6

⇒ g(x) = $x^{2} - 4x + 9$

⇒ g(6) = $(6)^{2} - 4 (6) -9$

           = 36 - 24 - 9

           = 3

g(5) = 3 which is a positive value

The root lies between the 5 and 6

Iteration:

Let $x_{1} = 5$  and $x_{2} = 6$

Let

$x_{3} = \frac{1}{2} (x_{1} + x_{2} )$

    = $\frac{1}{2} (5+6)$

    = $\frac{1}{2} (11)$

   = 5.5

g(5.5) = $(5.5)^{2} - 4(5.5) -9$

         = 30.25 - 22 - 9

         = -0.75

Therefore -0.75 < 0 which is negative

Final answer:

The required root lies between 5 and 6

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