Understanding the Relationship between Square Roots and Absolute Values

[khurram usman]

absolute value help ...

i am a bit confused about two things...
first: is sqrt(x^2) is the same thing as |x|? if yes then consider a negative no. 'z' ... i had a question like this: sqrt(z^2)...its answer was -z...so i suppose this means that this square root/ square and absolute value thing is the same and since z is negative so we say that answer is -z which will obviously be positive.

second: its related to the first one.
consider sqrt[-3]*sqrt[-3]
is this even computable in the real plane? i am asking this because the calculator was given a mth error...on the other hand alphawolfram resource was giving -3
i am with the resource on this because if my first question is correct then it is the same thing that is sqrt(-3)^2
so clarify all this

 

[Mentallic]

i am a bit confused about two things...
first: is sqrt(x^2) is the same thing as |x|? if yes then consider a negative no. 'z' ... i had a question like this: sqrt(z^2)...its answer was -z...so i suppose this means that this square root/ square and absolute value thing is the same and since z is negative so we say that answer is -z which will obviously be positive.

The definition of sqrt(x) is taking the principal root of x, which is just the positive value. It's different to solving the equation y²=x for y, because in that case we get $y=\pm\sqrt{x}$ which is both the positive and negative of sqrt(x).
So what this means is that if we take some number, then we square it, we've lost some information about the number - mainly whether it was positive or negative to begin with. For this reason, if we consider $\sqrt{x^2}$ since we need to square x first, then we take the square root again, we don't end up back at x exactly, unless x was positive. If x was negative then we've lost that information and that's why the answer is |x|.

second: its related to the first one.
consider sqrt[-3]*sqrt[-3]
is this even computable in the real plane? i am asking this because the calculator was given a mth error...on the other hand alphawolfram resource was giving -3
i am with the resource on this because if my first question is correct then it is the same thing that is sqrt(-3)^2
so clarify all this

This time it's different. We now have $$(\sqrt{x})^2$$ which isn't quite the same as $$\sqrt{x^2}$$
If we're allowed to work with the complex numbers, then $\sqrt{-3}=\sqrt{3}i$ where $i=\sqrt{-1},i^2=-1$ which is imaginary. In thise case, if we computed $$(\sqrt{-3})^2$$ we would end up with -3 because $(\sqrt{3}i)^2=(\sqrt{3})^2i^2=3(-1)=-3$

The reason your calculator gave a math error is because it can't work in the complex plane. Most handheld calculators have been designed that way.

 

[khurram usman]

The definition of sqrt(x) is taking the principal root of x, which is just the positive value. It's different to solving the equation y²=x for y, because in that case we get $y=\pm\sqrt{x}$ which is both the positive and negative of sqrt(x).
So what this means is that if we take some number, then we square it, we've lost some information about the number - mainly whether it was positive or negative to begin with. For this reason, if we consider $\sqrt{x^2}$ since we need to square x first, then we take the square root again, we don't end up back at x exactly, unless x was positive. If x was negative then we've lost that information and that's why the answer is |x|.

ok i understand the information lost thing...its a nice way of thinking of square roots.
the problem is that we again ended up with |x| which again will always be positive. so again if before squaring x was a negative no. then that information is lost.

is it necessary to do the whole iota procedure for the second part of my question? why don't we simply cancel out the square and square root?

 

[Mentallic]

the problem is that we again ended up with |x| which again will always be positive. so again if before squaring x was a negative no. then that information is lost.

Exactly

is it necessary to do the whole iota procedure for the second part of my question? why don't we simply cancel out the square and square root?

Ofcourse. I was just showing you that it works when dealing with complex numbers in the same way.

 

[khurram usman]

Exactly


Ofcourse. I was just showing you that it works when dealing with complex numbers in the same way.

thanku very much...understood!
by the way i have sent you a friend request...actually i am starting my calculus in college...so i will be needing help a lot and you explain very well
so do accept the request and once again thanks

 

[Mentallic]

Sure, but trust me that there are many other helpers on this forum that are very clear with their explanations as well