y^3 = ( 4 * ( x ^ 2 ) ) + 28
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salma omarii
on 8 Oct 2014
Commented: salma omarii
on 8 Oct 2014
I have solved the following by hand but am having difficulties implementing the code. If anyone is able to assist me I would great appreciate it.
I would like to use Newton-Raphson to solve:
y^3 = ( 4 * ( x ^ 2 ) ) + 28
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Accepted Answer
Sean de Wolski
on 8 Oct 2014
If you have symbolic math toolbox:
solve('y^3 = ( 4 * ( x ^ 2 ) ) + 28','x')
ans =
(y^3 - 28)^(1/2)/2
-(y^3 - 28)^(1/2)/2
solve('y^3 = ( 4 * ( x ^ 2 ) ) + 28','y')
ans =
4^(1/3)*(x^2 + 7)^(1/3)
4^(1/3)*((3^(1/2)*i)/2 - 1/2)*(x^2 + 7)^(1/3)
-4^(1/3)*((3^(1/2)*i)/2 + 1/2)*(x^2 + 7)^(1/3)
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