Topic: Engineering Mathematics
A real root of equation $x^{3}-5x-7=0$ by the method of false position correct to three decimal places is
- A. 2.747
- B. 2.084
- C. 2.077
- D. None of these
Answer: A. 2.747
Explanation: Let $f(x) = x³ - 5x - 7 = 0$.
$f(2) = 8 - 10 - 7 = -9$.
$f(3) = 27 - 15 - 7 = 5$.
The root lies between 2 and 3.
Using the false position method, $x_1 = \frac{a f(b) - b f(a)}{f(b) - f(a)} = \frac{2(5) - 3(-9)}{5 - (-9)} = \frac{10 + 27}{14} = 2.643$.
$f(2.643) = -1.752$. Since $f(2.643) < 0$, the root lies between 2.643 and 3.
Iterating further yields $x_2 \approx 2.736$, and eventually converges to $2.747$.