Can x be written as both a positive and negative value in an equation?

AI Thread Summary
The discussion clarifies that the square root function, denoted as √, only yields the positive root of a number, which is why the notation ±√ is often used in equations. It explains that for any real number x, √(x²) equals |x|, not x itself, highlighting that negative values are not returned by the square root function. An example with x = -2 illustrates that √(x²) results in a positive value. The conversation also notes that if additional information about x is known, such as it being positive, the simpler notation can be used. Understanding these distinctions is essential for correctly interpreting and solving equations involving square roots.
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<< Mentor Note -- OP has been advised to type their questions into the forum next time, instead of inserting images >>[/color]

1. Homework Statement

Dear Mentors and PF helpers,

I have a question from today's lesson.

image.jpg


Homework Equations

The Attempt at a Solution



Is 1) and 2) both acceptable? If only of of them is acceptable, is there any reason why?

Thanks for your time
 
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Only 2.
The ## \sqrt{} ## implies that it only returns the positive square root of a number. That is why you will often see ##\pm \sqrt{} ## in formulas.
 
For x any real number, \sqrt{x^2}= |x|, NOT \sqrt{x}= x

That is easily seen by taking x to be any negative number, say x= -2. In that case, x^2= (-2)^2= 4 and \sqrt{x^2}= \sqrt{4}= 2, NOT -2.
 
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However, if you know something about x, like it is positive, then you may use the notation in (1) .
Also, if you have ##y = x^2##, then ##x = \pm \sqrt{y}##.
 
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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