
#19
Nov112, 01:21 PM

P: 273

[tex]{ ({ { x }^{ 2 } }) }^{ \frac { 1 }{ 2 } }=x\\ \\ \sqrt { x } =\quad x\\ \sqrt { x } =\quad x[/tex]
How would I get x from using the exponent method? edit: Never mind. I'm an idiot. 



#20
Nov112, 02:42 PM

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$$ \sqrt { x^2 } = x $$ and similarly for the third line. 



#21
Nov112, 03:23 PM

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If the problem were to solve the equation, [itex]x^2= 4[/itex], then the correct answer would be [itex]\pm 2[/itex]. If the problem were to find [itex]\sqrt{4}[/itex] then the correct answer is [itex]2[/itex] NOT "[itex]\pm 2[/itex]". If the problem were to solve [itex]x^2= 5[/itex], then the correct answer is [itex]\pm\sqrt{5}[/itex]. Why do you need the "[itex]\pm[/itex]"? Because it is NOT part of the square root. IF [itex]\sqrt{5}[/itex] itself mean both positive and negative values you would NOT need the "[itex]\pm[/itex]". 



#22
Nov112, 03:25 PM

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The distinction I'm making is between the square root function (as denoted by the √ symbol), and the generic concept of a square root. The square roots of x^{2} are ±√(x^{2}) = ±x, which is the same as ±x. The point of disagreement is extremely subtle: the use of the definite article. I wrote the "square root" has two possible valuesi.e. in the generic sense of square root; you prefer to reserve "the square root" to mean the principal square root. Fair enough, but I think that's a matter of taste, and I'm not sure how else I could have worded it. "A square root has two possible vales"? No. "Taking the square root produces two possible values"? Still that definite article. 



#23
Nov112, 03:37 PM

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#24
Nov112, 04:28 PM

P: 273





#25
Nov112, 05:03 PM

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the disagreement really boils down to if you adhere to the oldfashioned idea of the "square root" as a polyvalued function, or adhere strictly to the idea that "the square root operation" is a singlevalued function.
The idea of polyvalued functions is really redundant, since you can define onevalued function to the power set of R, or any other set. but, I'm digressing here.. 



#26
Nov212, 02:21 AM

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#27
Nov212, 10:21 AM

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Later in that same post, haruspex said this: (x^{2})^{1/2} = ##\sqrt{x^2}## = ±x A reasonable inference from "does not define a function in the strict sense, so it returns a ± result" is that (x^{2})^{0.5} is multivalued. A graph of the equation y = (x^{2})^{1/2} shows that there is no x value that is mapped to more than one y value. 



#28
Nov312, 09:48 AM

P: 273

Can somebody confirm if I'm doing this right?
[tex]{ x }^{ 2 }=4\\ x=\sqrt { 4 } \\ x=\pm \sqrt { 4 } \\ x=\pm 2[/tex] [tex]{ { (x }^{ 2 }) }^{ 0.5 }=4\\ x=4\\ x=\pm 4[/tex] [tex]{ x }^{ 4 }={ 2 }^{ 4 }\\ { x }^{ 2 }=\sqrt { { 2 }^{ 4 } } \\ x=\sqrt { \sqrt { { 2 }^{ 4 } }  } [/tex] 



#29
Nov312, 01:09 PM

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In the first equation, there are two solutions. In the second equation, there is only one solution, since √4 is the same as √4, which is 2. x = 2 x = ±2 x^{4} = 2^{4} = 16 x^{4}  16 = 0 (x^{2}  4)(x^{2} + 4) = 0 (x  2)(x + 2)(x^{2} + 4) = 0 x = 2 or x = 2 There are also two imaginary solutions  x = 2i and x = 2i. 



#30
Nov312, 09:26 PM

P: 273

How did you go from x^{4}  16 = 0 to (x^{2}  4)(x^{2} + 4) = 0? What about the absolute value signs? I know it's a difference of square, but what if was 15 instead of 16?




#31
Nov312, 11:45 PM

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x^{4} = 15 x^{4}  15 = 0 (x^{2}  √15)(x^{2} + √15) = 0 (x  ## \sqrt[4]{15}##)(x + ## \sqrt[4]{15}##)(x^{2} + √15) = 0 x = ## ±\sqrt[4]{15}## There are also two imaginary roots. 



#32
Nov412, 12:42 AM

P: 685

I think the confusion stems from two things:
1. Solving for x in the equation x^2 = a. 2. Using the square root function. If you are given the equation x^2 = a, then there are two solutions, namely a positive and a negative: [itex]x = \sqrt{a}[/itex] and [itex]x = \sqrt{a}[/itex]. However, if you are using the square root function, i.e. [itex]\sqrt{x}[/itex], it only returns one value. To make this more clear, define a function [itex]s: X \rightarrow Y[/itex] [itex]s(x) = y[/itex], such that y>0 and y^2=x. This function s returns only one value for an input x. Replace s by the symbol [itex]\sqrt{}[/itex]. 



#33
Nov1112, 10:20 PM

P: 273

Wait a second... Sorry for my repeated lack of understanding, but how did I go from step 2 to step 3 here:
[tex]{ x }^{ 2 }=4\\ x=\sqrt { 4 } \\ x=\pm \sqrt { 4 } \\ x=\pm 2[/tex] If I remember correctly, to take the absolute value of anything, you just ensure that the sign is positive. For example, 2=2, 4=4, sqrt(4)=sqrt(4), and sqrt(4)=4. 



#34
Nov1212, 12:12 AM

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The difficulty isn't going from step 2 to step 3  it's going from step 1 to step 2.
x^{2} = 4 ## \Leftrightarrow## x^{2}  4 = 0 ## \Leftrightarrow## (x  2)(x + 2) = 0 ## \Leftrightarrow## x = ± 2 √4 = √4 = 2, not ±2. BTW, tahayassen, you had essentially the same set of equations in post #28, a week ago. I missed that you had gone from x^{2} = 4 to x = √4. The problem here is that the first equation has two solutions, while the second equation has only one. I have gone back and edited my post. 



#35
Nov1212, 12:18 AM

P: 34

[tex]{ x }^{ 2 }=4\\ \sqrt{{ x }^{ 2 }}=\sqrt { 4 } \\ x = 2 \\ x=\pm 2[/tex] 


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