Very simple concept is bugging me.

[shadoWizard]

Lets say we have Sqrt[(-3)^2]

 

if i solve it the way i know i'll get -3. But recently I was watching a math video online and it said that the solution would be +3. Why is that? I must add that the way I solve this expression is as follows.

http://www.www.urch.com/forums/attachments/gre-math/6611-very-simple-consept-bugging-me-untitled.jpg

 

Is this the wrong way of solving the square root?

 

Please help

 

 

Regards

Edited November 7, 2012 by shadoWizard

[lil2011]

I think You can not bring out -3 from radical unless you square it.

your way is correct and because of that when we have (9)1/2----> we have two root -3 and 3 because both make your equation correct.

(3)^2=9

(-3)^2=9

so (9)^(0.5)= 3 or -3

But you are not eligible for bringing out -3 below the radical. so (-3)^0.5 is not possible.you should square -3 below the radical and then bringing it out. this time you are sure the root was -3 not 3 because you have seen it before squaring it.

Best

[shadoWizard]

lil2011 thanks for the answer. But this is what i needed to ask that why am i not eligible to bring out -3 as you said?

[lil2011]

Its not defined - as you know the radical of negative number is imaginary

are you familiar with this kind of numbers: 3+3i

actually this is 3+3(-1)^(1/2)

we don't have this concept at GRE exam. its related to advanced math. so that you should only think about real number.

Best

[shadoWizard]

But when I multiply the 2 and 1/2 (in exponents) both the twos are cancelled out so we never touch -3. So before we reach the -ve number, sq. root is cancelled... isn't it?

[lil2011]

When you omit radical you obtain (1/2) they are not there simultaneously.So that you can not do that with negative number.negative number can not come out unless you say : first I square (-3) so it would be 9 below the radical then I square root from 9 which is +3 and come out.