Question

The root of the equation $x^{3} + x - 3 = 0$ lies in interval

(1, 2) after second iteration by false position method, it will be in

• A. (1.178, 2.00)
• B. (1.25, 1.75)
• C. (1.125, 1.375)
• D. (1.875, 2.00)

Answer 1

Answer: (A)

(1.178, 2.00)

Answer 2

$f(x) = x^{3} + x - 3$

$f(1) = - 1$ and $f(2) = 7$

Therefore, root lie in (1, 2).

Now, take $x_{0} = 1$, $x_{1} = 2$

$x_{2} = 1 - \frac{2 - 1}{+ 7 - ( - 1)}.( - 1)$ =1.125 and so $f(x_{2}) = - 0.451$

Hence, roots lie in (1.125, 2)

$\Rightarrow x_{3} = 1.125 - \frac{2 - 1.125}{7 - ( - 0.451)}( - 0.451) = 1.178$. So required root

lie in (1.178, 2)