find real root of x3-9x+1=0 by method of false position
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The third approximation of roots of x
3
−9x+1=0 in the interval (2,4) by the method of false position is?
Hard
Solution
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Correct option is
C
2.85
Here, x
3
−9x+1=0
Let f(x)=x
3
−9x+1
First Iteration:
Here, f(2)=−9<0 and f(4)=29>0
Now, Root lies between x
0
=2 and x
1
=4
x
2
=x
0
−f(x
0
)×
f(x
1
)−f(x
0
)
x
1
−x
0
=2−(−9)×
29−(−9)
4−2
=2.47368
Second Iteration:
Here, f(2.47368)=−6.1264 and f(2)=29>0
Now, Root lies between x
0
=2.47368 and x
1
=4
x
3
=x
0
−f(x
0
)×
f(x
1
)−f(x
0
)
x
1
−x
0
=2.47−(−6.13)×
29−(−6.13)
4−2.47
=2.73989
Third Iteration:
Here, f(2.73989)=−3.09067 and f(4)=29>0
Now, Root lies between x
0
=2.73989 and x
1
=4
x
4
=x
0
−f(x
0
)×
f(x
1
)−f(x
0
)
x
1
−x
0
=2.74−(−3.09)×
29−(−3.09)
4−2.74
=2.86125