Question

Find a real root of the equation x3-5x-3=0 by the method of false position correct to three decimal
places.

Answer 1

The real root of the equation is =2.747

Explanation:

• Since here range [a,b] is not given so by seeing option we will assume a=2 and b=3

f(x)=x³-5x-7

f(2)=(2)³-5(2)-7=-9

f(3)=(3)³-5(3)-7=5

f(2) f(3)<0

• There must be one root between 2 and 3

Now,

x1=af(b)-b(f(a))/f(b)-f(a)=2(5)-3×(-9)/5-(-9)

=10+27/14=2.643

f(x1)=(2.643)³-5(2.643)-7=-1.752<0

Now root between 2.643 and 3

x2=2.643×5-3(1.752)/5-(-1.752)=2.736

f(x)=(2.736)³-5(2.736)-7

=0.199<0

• Root lies between 2.736 and 3

x3=2.736×5-3(-0.199)/5-(-0.199)=2.746

f(3)=(2.746)³-(2.746)-7

=0.023<0

• The root lies between2.746 and 3

x4=2.746×5-3(-0.023)/5-(-0.023)

=2.747