ASKSAGE: Sage Q&A Forum - RSS feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Tue, 25 Jun 2013 10:00:31 +0200solving radical equations with parametershttps://ask.sagemath.org/question/10277/solving-radical-equations-with-parameters/I would like to find the solutions $y$ to this type of equations: $$\left(1+x -\sqrt{(1+x)^2-4y}\right)^2=z$$
with conditions on $x,y,z$ (like $0\lt y\lt x\leq \frac18$ and $0\lt z\lt x^2$).
Using `solve` with the option `to_poly_solve`:
sage: solve((1+x - sqrt((1+x)^2-4*y))^2 == z, y, to_poly_solve=True)
[y == 1/2*x^2 - 1/2*(x + 1)*sqrt(x^2 + 2*x - 4*y + 1) + x - 1/4*z + 1/2]
does not seem to work because $y$ appears on the right side of the solution. I expect to find a solution like
$$y=\frac14\left((1+x)^2-\left(1+x-\sqrt{z}\right)^2\right).$$
I also tried the same after specifying the conditions with `assume()`, without success.nejimbanTue, 25 Jun 2013 10:00:31 +0200https://ask.sagemath.org/question/10277/Cannot solve equation with two radical termshttps://ask.sagemath.org/question/8544/cannot-solve-equation-with-two-radical-terms/I am learning Sage on the Notebook by reworking examples in my old algebra book (starting with page one).
Could someone please explain the following behavior and how to solve the original equation?
This equation isn't getting solved:
solve(sqrt(2*x - 5) - sqrt(x - 3) == 1, x)
The output is:
[sqrt(x - 3) == sqrt(2*x - 5) - 1]
But the solution is x == 7 or x == 3
I tried the terms and they are solved:
solve(sqrt(2*x-5) == 1, x)
[x == 3]
solve(sqrt(x-3) == 1 , x)
[x == 4]
Thank youhunannerSun, 11 Dec 2011 09:10:35 +0100https://ask.sagemath.org/question/8544/Solving radical inequalitieshttps://ask.sagemath.org/question/8157/solving-radical-inequalities/Can this be solved in Sage?
> x-4>sqrt(x-2)
The standard solve method does not work, and neither does solve_ineq.Eviatar BachFri, 10 Jun 2011 04:15:16 +0200https://ask.sagemath.org/question/8157/