Question

One who solved my 4th iteration, please sir help me in finding 5th iteration too. First Iteration: x0 + x^2 + 25 = 225x^2 = 25, x^2 = 2 f(x) = -3f(x) = 3125f(x) = 2253 - 3225 - 503600 Since f(x) is negative, therefore the root lies between 225 and 25. Since f(x) is negative, therefore x becomes new x and x remains constant. Second Iteration: x = 225 x = 25 y = x^2 - 03600 f(x) = 3125 - 1225 + 25 - 2375^2 f(x) = 2375 - 32375 - 512714 Since f(x) is positive, therefore the root lies between 225 and 2375. Since f(x) is positive, therefore x becomes new x and x remains the same. Third Iteration: x = 225 x = 2375 y = x^2 + x^2 - 3125 f(x) = -03600 f(x) = 12714 f(x) = 04289 Since f(x) is positive, therefore x becomes new x and x remains the same. Fourth Iteration:

          One who solved my 4th iteration, please sir help me in finding 5th iteration too.

First Iteration:
x0 + x^2 + 25 = 225x^2 = 25, x^2 = 2
f(x) = -3f(x) = 3125f(x) = 2253 - 3225 - 503600

Since f(x) is negative, therefore the root lies between 225 and 25. Since f(x) is negative, therefore x becomes new x and x remains constant.

Second Iteration:
x = 225
x = 25
y = x^2 - 03600
f(x) = 3125 - 1225 + 25 - 2375^2
f(x) = 2375 - 32375 - 512714

Since f(x) is positive, therefore the root lies between 225 and 2375. Since f(x) is positive, therefore x becomes new x and x remains the same.

Third Iteration:
x = 225
x = 2375
y = x^2 + x^2 - 3125
f(x) = -03600
f(x) = 12714
f(x) = 04289

Since f(x) is positive, therefore x becomes new x and x remains the same.

Fourth Iteration:

one who solved my 4th iterationplease sir help me in finding 5th iteration too first iteration xox 225 225 x2x25x 2 2 fx 3fx3125fx2253 3225 5 03600 since fx is negativetherefore the root lie 02559

Added by Juan W.

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