ASKSAGE: Sage Q&A Forum - Individual question feedhttps://ask.sagemath.org/questions/Q&A Forum for SageenCopyright Sage, 2010. Some rights reserved under creative commons license.Sun, 11 Dec 2011 09:00:49 -0600Cannot 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 youSun, 11 Dec 2011 02:10:35 -0600https://ask.sagemath.org/question/8544/cannot-solve-equation-with-two-radical-terms/Answer by benjaminfjones for <p>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?</p>
<p>This equation isn't getting solved:</p>
<pre><code>solve(sqrt(2*x - 5) - sqrt(x - 3) == 1, x)
</code></pre>
<p>The output is:</p>
<pre><code>[sqrt(x - 3) == sqrt(2*x - 5) - 1]
</code></pre>
<p>But the solution is x == 7 or x == 3</p>
<p>I tried the terms and they are solved:</p>
<pre><code>solve(sqrt(2*x-5) == 1, x)
[x == 3]
solve(sqrt(x-3) == 1 , x)
[x == 4]
</code></pre>
<p>Thank you</p>
https://ask.sagemath.org/question/8544/cannot-solve-equation-with-two-radical-terms/?answer=12993#post-id-12993Try 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-resultSun, 11 Dec 2011 09:00:49 -0600https://ask.sagemath.org/question/8544/cannot-solve-equation-with-two-radical-terms/?answer=12993#post-id-12993Answer by hunanner for <p>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?</p>
<p>This equation isn't getting solved:</p>
<pre><code>solve(sqrt(2*x - 5) - sqrt(x - 3) == 1, x)
</code></pre>
<p>The output is:</p>
<pre><code>[sqrt(x - 3) == sqrt(2*x - 5) - 1]
</code></pre>
<p>But the solution is x == 7 or x == 3</p>
<p>I tried the terms and they are solved:</p>
<pre><code>solve(sqrt(2*x-5) == 1, x)
[x == 3]
solve(sqrt(x-3) == 1 , x)
[x == 4]
</code></pre>
<p>Thank you</p>
https://ask.sagemath.org/question/8544/cannot-solve-equation-with-two-radical-terms/?answer=12992#post-id-12992I 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.Sun, 11 Dec 2011 02:21:40 -0600https://ask.sagemath.org/question/8544/cannot-solve-equation-with-two-radical-terms/?answer=12992#post-id-12992