I was a little confused about that simplification in a diferential equation, but thanks for the replyMaybe what you are asking is why the claim in your title, not the example in the question body, is false, as reported to you by someone else. If so, they are right. The example is not general.
That's because, technically, sqrt(x^2+x^2*y^2) is equal to |x|(sqrt(1+y^2)):
sqrt(x^2+x^2*y^2) = sqrt(x^2(1 +y^2)) = sqrt(x^2)sqrt(1 +y^2) = |x|sqrt(1 +y^2)
The absolute value is necessary if x can be negative.
I don't appreciate the implied threat in your user name.
Not true, you do have a problem as Dr Peterson said that he does not like your user name. Since this is a free service you should consider changing your username out of respect to one of the main helpers on the forum.And dont worry about the name I've been using it for years and have not problems, much regards!
Technically, as I realized after I said that, "your dead" is not the threat that "you're dead" would be; it just says that he is our dead person. Not exactly life-affirming, but not a threat. I wouldn't use it myself ...Not true, you do have a problem as Dr Peterson said that he does not like your user name. Since this is a free service you should consider changing your username out of respect to one of the main helpers on the forum.
See post #2. I believe the question was about why it is not true in general that sqrt(x^2+x^2*y^2) = x sqrt(1 +y^2), while (as you confirmed) it is true for the example, where x = 2.sqrt(2^2+3^2*2^2)
=sqrt( 2^2(1+3^2))
=sqrt( 2^2) sqrt(1+3^2)
=2* sqrt(1+3^2)
It is true