Why can you take the positive only?

[kenhung123]

If I have an equation like this It actually works if I root both sides like But how come I don't need to

[Aqualim]

If I have an equation like this It actually works if I root both sides like But how come I don't need to

Lol

I'm thinking maybe because zero doesn't change no matter whether it is positive or negative?

[TrueTears]

u cant square root both sides...

sqrt[f(x)+g(x)] =/= sqrt[f(x)g(x)]

[kamil9876]

if

then

 OR

which is

  OR  

but so it just happens that the second case does not happen. Of course if the question instead was to solve:



then it turns out that the negative root gives you the money, but the positive doesn't. So take all roots always and eliminate a possiblity when you can see it.

[kamil9876]

u cant square root both sides...

sqrt[f(x)+g(x)] =/= sqrt[f(x)g(x)]

I gave him the benefit of the doubt that he rearanged first

[kenhung123]

Wait is it because so you can have OR is this just what you did lol???

[TrueTears]

u cant square root both sides...

sqrt[f(x)+g(x)] =/= sqrt[f(x)g(x)]

I gave him the benefit of the doubt that he rearanged first

oh i see the wording was weird though.

[stonecold]

Isn't it because nothing has actually been squared, so you know what the originals are, so when you square root, you just keep the originals?

[TrueTears]

id solve the Q like this:

(x-1)^2 = (4-x)^2

|x-1| = |4-x|

|(x-1)/(4-x)|=0

solve for x this should be simple.

it's nothing to do with taking the positive only, split into cases, and solve it.

[kamil9876]

That's really overcomplicating it and making it harder.

I was just trying to answer a general question about how to solve equations and what "takings something of both sides" really means.

As for the question specifically it is also good (apart from what we did earlier which is fine and "has something to do with the question") to just use different of perfect squares:



so x=2/5

[TrueTears]

nah its alright, a systematic way of solving it, follow the steps and you cant go wrong.

[m@tty]

id solve the Q like this:

(x-1)^2 = (4-x)^2

|x-1| = |4-x|

|(x-1)/(4-x)|=0

solve for x this should be simple.

it's nothing to do with taking the positive only, split into cases, and solve it.

You mean |(x-1)/(4-x)|=1 , yeah?

[TrueTears]

yeah 1, bad mistake.

[kamil9876]

The first approach was the most obvious, I don't see how absolute values make things simpler.

Try

Using your method.

[TrueTears]

no ur method was a consequence of the modulus, except you just skipped it, you picked 2 certain cases, its just as easy to not leave out steps, the other cases are just redundant as they not satisfy the domain, so leaving it in only creates a fuller picture. this is quite trivial though, nothing too flashy about this Q.