Question

In which interval does the root of equation x^3-2x-5=0 lie? Explain how to solve it.

Answer 1

Let f(x) = x³ - 2x - 5
Now, choose two points a and b randomly in such a way f(a) > 0 /f(b) < 0 and vice -versa .
Let we choose 2 and 3
then, f(2) = (2)³ - 2(2) - 5 = 8 - 4 - 5 = -1
f(3) = (3)³ - 2(3) - 5 = 27 - 6 - 5 = 16
Here , we see f(2) < 0 and f(3) > 0 , it means one root of f(x) lies between 2 and 3

For more precise , take average of 2 and 3 e.g., (2 + 3)/2 = 2.5
Now, put in f(x)
f(2.5) = (2.5)³ - 2(2.5) - 5
= 15.625 - 5 - 5 = 15.625 - 10 = 5.625 >0 , hence you can say one root of f(x) lies between 2 and 2.5 . If you want more precise then again do average between 1 and 2.5 and find f(x) , if f(x) in that value is positive then root lies between 2 and average of 1 and 2.5 . For more precise again and again try and solve .