Question

The equation \( x^3 - 6x - 4 = 0 \) has one real root between 2 and 3. The 2ndapproximation to the root of the equation by bisection method will be

• A. 2.25
• B. 2.53
• C. 2.75
• D. 2.57

Hint:

Always check signs of function values to update interval in each bisection step.

Answer 1

The Correct Option is C

Let \( f(x) = x^3 - 6x - 4 \). We apply the bisection method starting with interval [2, 3]:
\[ f(2) = 8 - 12 - 4 = -8,\quad f(3) = 27 - 18 - 4 = 5 \]
Since \( f(2) \cdot f(3) < 0 \), a root lies in [2, 3].
1st Approximation: Midpoint \( x_1 = \frac{2 + 3}{2} = 2.5 \),
\[ f(2.5) = (2.5)^3 - 6(2.5) - 4 = 15.625 - 15 - 4 = -3.375 \]
Now new interval becomes [2.5, 3] (since sign change between f(2.5) and f(3)).
2nd Approximation: \( x_2 = \frac{2.5 + 3}{2} = 2.75 \)