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

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Topic review (newest first)
2013-07-26 20:50:51

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

2013-07-19 03:03:20


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

2013-07-19 02:54:36

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?

2013-07-17 07:45:21

That is only two distinct ways.

2013-07-17 07:17:12

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

2013-07-17 07:08:46

Hi atran;

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

(x+3)^2 = 13

(x+3) = √13

2013-07-17 06:56:38

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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