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#1 2013-07-16 08:56:38

atran
Member
Registered: 2013-07-12
Posts: 91

Why is the square root of (x+y)^2 not (-x-y)?

Hi again, this question may seem silly, but I'm confused.

Say, (x^2)+6x-4=0, then by completing the square I get:
(x^2)+6x-4= 0
(x^2)+6x   = 4
(x^2)+6x+9= 4+9
(x+3)^2= 13
Now, why isn't sqrt((x+3)^2) also equal to -(x+3)=-x-3?

Many small questions have been popping up in my head. This is leading me to a confused state. I used to do well and understand algebra, but I don't what happened, things started becoming confusing and unclear.

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#2 2013-07-16 09:08:46

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,237

Re: Why is the square root of (x+y)^2 not (-x-y)?

Hi atran;

It is usually written like this but you are essentially correct.

(x+3)^2 = 13

(x+3) = ±√13


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#3 2013-07-16 09:17:12

atran
Member
Registered: 2013-07-12
Posts: 91

Re: Why is the square root of (x+y)^2 not (-x-y)?

So there are four ways of expressing it?

1) x+3=√13 => x=√(13)-3
2) -(x+3)=-√13 => x=√(13)-3
3) x+3=-13 => x=-√(13)-3
4) -(x+3)=√13 => x=-√(13)-3

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#4 2013-07-16 09:45:21

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,237

Re: Why is the square root of (x+y)^2 not (-x-y)?

That is only two distinct ways.


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#5 2013-07-18 04:54:36

atran
Member
Registered: 2013-07-12
Posts: 91

Re: Why is the square root of (x+y)^2 not (-x-y)?

I still don't get it. I learned that √(x) or x^(1/2) is the principal square root of x.

How to think when getting from this step [(x+3)^2 = 13] to [(x+3) = ±√13]?
I mean, why the last step is written like that? Why not (±(x+3)=√13)?
What makes both (±(x+3)=±√13) and (∓(x+3)=±√13) valid?

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#6 2013-07-18 05:03:20

bobbym
Administrator
From: Bumpkinland
Registered: 2009-04-12
Posts: 87,237

Re: Why is the square root of (x+y)^2 not (-x-y)?

Hi;

I still don't get it. I learned that √(x) or x^(1/2) is the principal square root of x.

Who says you should only use the principal root in this case. That would only get one root, a quadratic has 2 roots. See post #2


In mathematics, you don't understand things. You just get used to them.
Of course that result can be rigorously obtained, but who cares?
Combinatorics is Algebra and Algebra is Combinatorics.

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#7 2013-07-25 22:50:51

math.guru.92
Member
Registered: 2013-07-25
Posts: 3

Re: Why is the square root of (x+y)^2 not (-x-y)?

It is simple. Here we are dealing with the polynomial of degree "2" so when we will take a an under root on both sides, we will find two solutions of x one with a positive sign and one with negative sign.
(x+3)^2 = 13  ---------- original
when we take under root we get two equations
(x+3) = +√13 ----(equation 1)                                 (x+3) = -√13 ------- (equation 2)
equation 1 goes to form                                           equation 2 goes to form
x=+√13 - 3                                                                x=-√13 -3

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