Question

find the root of the equation x³-2x²-5=0 by bisection method​

Answer 1

Answer:

I have not learned that yet

Answer 2

Answer :-

The root of the equation upto 4 places of decimal = 2.0625

Solution :-

the equation is  x³ - 2·x - 5 = 0

Let f(x) = x³ - 2·x - 5

Now,

x       0              1          2          3

f(x)     -5           -6          -1         16

1st iteration :

Here f(2) = -1 < 0 and f(3) = 16 > 0

∴ Root lies between 2 and 3

$x_0=\frac{2+3}{2} =2.5$

$f(x_0)=f(2.5)= (2.5) \ . \ 3-2 \ . \ (2.5)-5= \ \ \ 5.625>0$

2nd iteration :

Here f(2) = -1 < 0 and f(2.5) = 5.625 > 0

∴ Now, Root lies between 2 and 2.5

$x_1=\frac{2+2.5}{2} =2.25$

$f(x_1)=f(2.25)= (2.25) \ . \ 3-2 \ . \ (2.25)-5= \ \ \ 1.89062>0$

3rd iteration :

Here f(2) = -1 < 0 and f(2.25) = 1.89062 > 0

∴ Now, Root lies between 2 and 2.25

$x_2=\frac{2+2.25}{2} =2.125$

$f(x_2)=f(2.125)= (2.125) \ . \ 3-2 \ . \ (2.125)-5= \ \ \ 0.3457>0$

4th iteration :

Here f(2) = -1 < 0 and f(2.125) = 0.3457 > 0

∴ Now, Root lies between 2 and 2.125

$x_3=\frac{2+2.125}{2} =2.0625$

$f(x_3)=f(2.0625)= (2.0625) \ . \ 3-2 \ . \ (2.0625)-5= \ \ \ -0.35132>0$

o, the root of the given equation upto 4 places of decimal is 2.0625

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