Is there a better way to find the roots of a quadratic equation?

[trap]

Let f(x)=x^2 - a, where a>0. The roots of the equation f(x)=0 are +root(a), -root(a).

Show that if x1 > 0 is any initial estimate for root(a), then the Newton-Raphson method gives the iteration formula

x n-1 = 1/2( xn + a/xn ),

n>=1

 

[trap]

Sorry, the last post wasn't clear, the iteration formula is..

Xn-1 = (1/2)(Xn + a/Xn) , n >= 1

 

[learningphysics]

Sorry, the last post wasn't clear, the iteration formula is..

Xn-1 = (1/2)(Xn + a/Xn) , n >= 1

Shouldn't this be:
Xn+1 = (1/2)(Xn + a/Xn) , n >= 1

Did you apply the Newton-raphson method Xn+1=Xn-f(xn)/f'(xn) ?

 

[trap]

yes, it's Xn-1 instead, my fault
I tried the Newton's method, but still have trouble solving the question

 

[learningphysics]

yes, it's Xn-1 instead, my fault
I tried the Newton's method, but still have trouble solving the question

Can you show what you did? The given formula arrives almost immediately by applying the method.

 

[Integral]

Newtons Method is;

x_{n+1} = x_n - \frac {f(x_n)} {f'(x_n)}

You are given

f(x)= x^2 - a

Compute f'(x) = 2x

Now apply Newtons method:
x_{n+1}= x_{n}- \frac {f(x_n)} {f'(x_n)} = x_n - \frac {x_n^2 - a } {2x_n} = \frac { 2 x^2_n - x^2_n +a} {2x_n}= \frac {x^2_n + a} {2x_n} = \frac 1 2 ( x_n + \frac a {x_n})

 

[trap]

thx for the help guys!

 

[trap]

kataKonik, hi..you must be taking the same course as i am...

 

[trap]

well...I'm not even sure if i can use the web to help me with these problem sets..so i just want to keep my identity as a secret