# GATE2010-2 [closed]

2 votes
1.3k views

Newton-Raphson method is used to compute a root of the equation $x^2 - 13 = 0$ with 3.5 as the initial value. The approximation after one iteration is

1. 3.575
2. 3.676
3. 3.667
4. 3.607
closed with the note: out of syllabus

closed
0

follow this one...

## 1 Answer

3 votes
x=3.5 - ((3.5^2)-13)/(2*3.5) so x=3.607 it is from the next term formula of newton raphson method
0
What is the formula, Sir?
0
0

I have seen this formula, but it looks like  Bhagirathi  is using some other formula

0
same formula is used, x1= x0-f(x0)/f'(x0)
Answer:

## Related questions

6 votes
1 answer
1
703 views
In the Newton-Raphson method, an initial guess of $x_0= 2$ is made and the sequence $x_0,x_1,x_2\:\dots$ is obtained for the function $0.75x^3-2x^2-2x+4=0$ Consider the statements $x_3\:=\:0$ The method converges to a solution in a finite number of iterations. Which of the following is TRUE? Only I Only II Both I and II Neither I nor II
0 votes
0 answers
2
283 views
Which of the following statements is true in respect of the convergence of the Newton-Rephson procedure? It converges always under all circumstances. It does not converge to a tool where the second differential coefficient changes sign. It does not converge to a root where the second differential coefficient vanishes. None of the above.
4 votes
2 answers
3
643 views
Newton-Raphson iteration formula for finding $\sqrt[3]{c}$, where $c > 0$ is $x_{n+1}=\frac{2x_n^3 + \sqrt[3]{c}}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 - \sqrt[3]{c}}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 + c}{3x_n^2}$ $x_{n+1}=\frac{2x_n^3 - c}{3x_n^2}$
0 votes
3 answers
4
760 views
The iteration formula to find the square root of a positive real number $b$ using the Newton Raphson method is $x_{k+1} = 3(x_k+b)/2x_k$ $x_{k+1} = (x_{k}^2+b)/2x_k$ $x_{k+1} = x_k-2x_k/\left(x^2_k+b\right)$ None of the above