Square-Root of a binomial squared

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The discussion centers on the correct assignment of the ± symbol in the equation derived from the square root of a binomial squared. It concludes that when solving the equation, the solutions can be expressed as 2x = ±(a-b), leading to two distinct cases: a = ±b or ±a = b. The conversation highlights that while some introductory algebra texts emphasize that √(u²) = |u|, the ± notation serves as a shortcut for identifying possible solutions. The participants express a desire for a more formal understanding of the implications of these cases. Overall, the thread clarifies the relationship between the ± notation and the solutions derived from squaring both sides of an equation.
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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?
 
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The solutions to that equation are simply ##2x = \pm (a-b)##.
 
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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 said:
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.
 
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PeroK said:
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!
 
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I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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