# Cannot solve equation with two radical terms

 0 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 you asked Dec 11 '11 This post is a wiki. Anyone with karma >150 is welcome to improve it. Anonymous

 3 Try using the to_poly_solve=True option. It's been on several people's wishlist to improve the documentation for the global solve function. sage: solve(sqrt(2*x - 5) - sqrt(x - 3) == 1, x, to_poly_solve=True) [x == 3, x == 7]  See also the answer to this older question: http://ask.sagemath.org/question/397/unexpected-solve-result answered Dec 11 '11 This post is a wiki. Anyone with karma >150 is welcome to improve it. benjaminfjones 2545 ● 4 ● 36 ● 67 http://bfj7.com/
 0 I saw a very similar question titled: "strange behaviour when solving equations symbolically" The answer basically was it can't do it. I guess I have been so impressed by Sage so far, that I thought it looked like a simple problem. answered Dec 11 '11 This post is a wiki. Anyone with karma >150 is welcome to improve it.

[hide preview]