Question

By bisection method, the real root of the equation $x^{3} - 9x + 1 = 0$ lying between $x = 2$ and $x = 4$ is nearer to

• A. 2.2
• B. 2.75
• C. 3.5
• D. 4.0

Answer 1

Answer: (B)

2.75

Answer 2

Since $f(2) = 2^{3} - 9(2) + 1 < 0$ and $f(4) = 4^{3} - 9(4) + 1 > 0$

$\therefore$ Root will lie between 2 and 4.

At $x = \frac{2 + 4}{2} = 3$, $f(3) = 3^{3} - 9(3) + 1 > 0$

$\therefore$ Root lie between 2 and 3.

At $x = \frac{2 + 3}{2} = 2.5$,$f(2.5)$ is –ive,$\therefore$ root lies between 2.5 and 3

At $x = \frac{2.5 + 3}{2} = 2.75$ and $f(2.75) = (2.75)^{3} - 9(2.75) + 1 < 0$.

$\therefore$Root is near to 2.75.