# Square-Root of a binomial squared

• lightlightsup
In summary, the equation has two solutions, depending on whether the number is greater than or less than the given integer, 3.f

#### lightlightsup

Homework Statement
##\sqrt{(a-b)^2}=\sqrt{4x^2}##
##±(a-b)=±2x##
Is this a correct distribution of the ##±##s?
Relevant Equations
None.
Suppose that ##a##, ##b##, and ##x## are integers.
How would the ##±##s be correctly assigned in such an equation?

The solutions to that equation are simply ##2x = \pm (a-b)##.

• lightlightsup and SammyS
It really doesn't matter. If we were to accept $$\pm a= \pm b$$ literally there would be four cases:

1) taking "+" on both sides, a= b
2) taking "-" on both sides, -a= -b
But this is just a= b with both sides multiplied by -1. They are the same.
3) taking "+" on the left side and "-" on the right, a= -b
4) taking "-" on the left side and "+" on the right, -a= b
But, again, the second is the same as the first, just with both sides multiplied by -1.

So there are really just two different cases which can be given by either ##a= \pm b## or ##\pm a= b##.

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• lightlightsup
Homework Statement:: ##\sqrt{(a-b)^2}=\sqrt{4x^2}##
##±(a-b)=±2x##
Is this a correct distribution of the ##±##s?
Relevant Equations:: None.

Suppose that ##a##, ##b##, and ##x## are integers.
How would the ##±##s be correctly assigned in such an equation?
Most introductory algebra textbooks (at least those of which I am aware) point out that actually, ##\sqrt{u^2 ~ }=|u| ## .

So for instance, solving an equation such as, ## t^2=9 ##, by taking the square root of both sides gives:
## \sqrt{t^2\ }=\sqrt{9\ } ##

simplifying gives: ##|t|=3 ## .
So that if ## t > 0 ##, then ## |t|=t## thus we have ##t = 3## .
However, if ## t < 0 ##, then ## |t|=-t## thus we have ##-t = 3## .

This gives the solutions: ##t = 3## or ##t=-3## .

I tend to think of the ##\pm## method as a short cut of sorts.

• lightlightsup and PeroK
The solutions to that equation are simply ##2x = \pm (a-b)##.
I deduced the same thing by just guessing that since both sides being positive or negative would lead to the positives/negatives cancelling out, then, the only remaining scenarios are:
one where only 1 negative sign, and
one where both sides are positive.

I was hoping for a more formal understanding of what was going on.

You guys have provided that to me.
Thank You!

• jim mcnamara