The question belong to engineering first year numerical methods
in Other Math Topics by

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
To avoid this verification in future, please log in or register.

1 Answer

This is a cubic, not a quadratic equation. 

The bisection method finds two values of x that make f(x)<0 and f(x)>0, then we change x so that it's the average of both attempts. The iterative method allows us to find such values for x.

When f(2)=-1 and f(3)=16 so there is a root between x=2 and 3, so now put x=2.5:

f(2.5)=5.625 so now we average 2 and 2.5=2.25. f(2.25)=1.89 approx, so we average 2 and 2.25=2.125.

f(2.125)>0 so we continue in the same way, finally arriving at x=2.0945 approx.

Another initial approach is to draw the graph of f(x) and note that it intersects the x-axis close to 2.1. This enables you to see what initial values of x to use.

by Top Rated User (1.0m points)

Related questions

1 answer
asked Feb 18, 2021 in Other Math Topics by anonymous | 207 views
1 answer
asked Feb 18, 2021 in Other Math Topics by anonymous | 201 views
1 answer
1 answer
asked Feb 18, 2021 in Other Math Topics by anonymous | 153 views
1 answer
asked Feb 18, 2021 in Other Math Topics by anonymous | 161 views
1 answer
asked Feb 18, 2021 in Other Math Topics by anonymous | 135 views
1 answer
asked Feb 18, 2021 in Other Math Topics by anonymous | 460 views
1 answer
asked Feb 18, 2021 in Other Math Topics by anonymous | 510 views
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
87,184 questions
97,329 answers
2,380 comments
24,564 users