Forum
»
Math question (Reps)
- Results 1 to 16 of 16
-
Page 1 of 1
Thread: Math question (Reps)
10-16-2020, 09:25 AM
-
#1
- anonkunbrah
- Join Date: Jul 2015
- Posts: 52,540
- Subscribers: 1
- Rep Power: 374460
-
-
Math question (Reps)
When you graph the following function you get this:

But when you square both sides and then move y^2 to the left, you get this:

Why does the first equation get a half circle and the 2nd equation get a full circle when both equations are the same (just in a different format). Sorry if this seems basic but sometimes it's the basics that trip me up

But when you square both sides and then move y^2 to the left, you get this:

Why does the first equation get a half circle and the 2nd equation get a full circle when both equations are the same (just in a different format). Sorry if this seems basic but sometimes it's the basics that trip me up
10-16-2020, 09:29 AM
-
#2
- BulkingIsHard
- Registered Bigot
-
- BulkingIsHard
- Registered Bigot
- Join Date: Aug 2016
- Posts: 12,670
- Rep Power: 70015
-
-
The second one isn't a function
https://en.wikipedia.org/wiki/American_decline
https://en.wikipedia.org/wiki/Societal_collapse#By_absorption
10-16-2020, 09:31 AM
-
#3
- anonkunbrah
- Join Date: Jul 2015
- Posts: 52,540
- Subscribers: 1
- Rep Power: 374460
-
-
Originally Posted By BulkingIsHard⏩
Could you elaborate?The second one isn't a function
10-16-2020, 09:33 AM
-
#4
- BulkingIsHard
- Registered Bigot
-
- BulkingIsHard
- Registered Bigot
- Join Date: Aug 2016
- Posts: 12,670
- Rep Power: 70015
-
-
Originally Posted By anonkunbrah⏩
Wikipedia:Could you elaborate?
In mathematics, a function is a binary relation between two sets that associates every element of the first set to exactly one element of the second set. Typical examples are functions from integers to integers, or from the real numbers to real numbers.

https://en.wikipedia.org/wiki/American_decline
https://en.wikipedia.org/wiki/Societal_collapse#By_absorption
10-16-2020, 09:35 AM
-
#5
- SniXSniPe
- Registered User
-
- SniXSniPe
- Registered User
- Join Date: Nov 2008
- Location: United States
- Posts: 2,326
- Rep Power: 5201
-
-
Originally Posted By anonkunbrah⏩
Think about it like this:When you graph the following function you get this:

But when you square both sides and then move y^2 to the left, you get this:

Why does the first equation get a half circle and the 2nd equation get a full circle when both equations are the same (just in a different format). Sorry if this seems basic but sometimes it's the basics that trip me up

But when you square both sides and then move y^2 to the left, you get this:

Why does the first equation get a half circle and the 2nd equation get a full circle when both equations are the same (just in a different format). Sorry if this seems basic but sometimes it's the basics that trip me up
Plot the points that are possible given the formula.
In the first formula, what values of X are possible?
For example, when X = 2, that means Y = 0.
10-16-2020, 09:37 AM
-
#6
- anonkunbrah
- Join Date: Jul 2015
- Posts: 52,540
- Subscribers: 1
- Rep Power: 374460
-
-
Originally Posted By BulkingIsHard⏩
Didn't really help me get it but repped anywaysWikipedia:
Originally Posted By SniXSniPe⏩
Actually i did that and i can see why it made the graph of a half circle. I think what threw me off is that doing something to both sides and rearranging an equation is supposed to result in the same thing. Yet doing that gets a different graphThink about it like this:
Plot the points that are possible given the formula.
In the first formula, what values of X are possible?
Plot the points that are possible given the formula.
In the first formula, what values of X are possible?
10-16-2020, 09:40 AM
-
#7
In the first:
If you discount imaginary numbers (j), the values of X can only be positive and the value of y cannot exceed 2 or be below -2 or else you end up with the squareroot of a negative number (and hence an imaginary number which can't be plotted here)
In the second:
both x and y can be positive or negative as you are squaring both but again neither can exceed +-2
If you discount imaginary numbers (j), the values of X can only be positive and the value of y cannot exceed 2 or be below -2 or else you end up with the squareroot of a negative number (and hence an imaginary number which can't be plotted here)
In the second:
both x and y can be positive or negative as you are squaring both but again neither can exceed +-2
Somehow still managing to avoid getting 'too big'
Non-CEO, 0.1235K per day
10-16-2020, 09:41 AM
-
#8
- reinsdorfsucks
- \_(ツ)_/
-
- reinsdorfsucks
- \_(ツ)_/
- Join Date: Sep 2009
- Location: Chicago, Illinois, United States
- Age: 37
- Posts: 4,084
- Rep Power: 29224
-
-
Edit: Wizard ^^^
I think because in the first one x can never be a negative number as a result of a square root, unless it's imaginary: -1^1/2
I think because in the first one x can never be a negative number as a result of a square root, unless it's imaginary: -1^1/2
Pureblood
10-16-2020, 09:41 AM
-
#9
10-16-2020, 09:42 AM
-
#10
Originally Posted By AlexSays⏩
this pretty muchIn the first:
If you discount imaginary numbers (j), the values of X can only be positive and the value of y cannot exceed 2 or be below -2 or else you end up with the squareroot of a negative number (and hence an imaginary number which can't be plotted here)
In the second:
both x and y can be positive or negative as you are squaring both but again neither can exceed +-2
If you discount imaginary numbers (j), the values of X can only be positive and the value of y cannot exceed 2 or be below -2 or else you end up with the squareroot of a negative number (and hence an imaginary number which can't be plotted here)
In the second:
both x and y can be positive or negative as you are squaring both but again neither can exceed +-2
You guys, I'm seriously
10-16-2020, 09:44 AM
-
#11
- anonkunbrah
- Join Date: Jul 2015
- Posts: 52,540
- Subscribers: 1
- Rep Power: 374460
-
-
Originally Posted By AlexSays⏩
so it's because the square root is out of the picture that the restriction lets up and thus will allow for negative values of x? i was baffled since for everything else, it gets the same graph between function rearrangement (y=x-2 is same as x=y+2)In the first:
If you discount imaginary numbers (j), the values of X can only be positive and the value of y cannot exceed 2 or be below -2 or else you end up with the squareroot of a negative number (and hence an imaginary number which can't be plotted here)
In the second:
both x and y can be positive or negative as you are squaring both but again neither can exceed +-2
If you discount imaginary numbers (j), the values of X can only be positive and the value of y cannot exceed 2 or be below -2 or else you end up with the squareroot of a negative number (and hence an imaginary number which can't be plotted here)
In the second:
both x and y can be positive or negative as you are squaring both but again neither can exceed +-2
10-16-2020, 10:22 AM
-
#12
- numberguy12
- Registered User
-
- numberguy12
- Registered User
- Join Date: Jan 2017
- Posts: 6,938
- Rep Power: 51915
-
-
The symbol for square root √ in mathematics is by convention the “principal root”....the nonnegative root.
Thus √(4) as written = 2. Not -2, or “plus or minus 2”......but 2. By convention.
So to answer your question.
The point (-2,0) does not satisfy your first equation x = √(4-y^2). It would be -2 = √(4-0^2)....but -2 cannot be the principal root of 4 (it’s negative).
The point (-2,0), however does satisfy the second equation x^2+y^2 = 4, and so is included in the graph.
If we actually solve for x in x^2+y^2 = 4 we get both halves x= √(4-y^2) or x= -√(4-y^2).
No need to bring up imaginary numbers etc, like some are doing above.
Thus √(4) as written = 2. Not -2, or “plus or minus 2”......but 2. By convention.
So to answer your question.
The point (-2,0) does not satisfy your first equation x = √(4-y^2). It would be -2 = √(4-0^2)....but -2 cannot be the principal root of 4 (it’s negative).
The point (-2,0), however does satisfy the second equation x^2+y^2 = 4, and so is included in the graph.
If we actually solve for x in x^2+y^2 = 4 we get both halves x= √(4-y^2) or x= -√(4-y^2).
No need to bring up imaginary numbers etc, like some are doing above.
∫∫ Mathematics crew ∑∑
♫1:2:3:4 Pythagoras crew ♫ ♫ 🧮
Nullius in verba
10-16-2020, 11:29 AM
-
#13
- numberguy12
- Registered User
-
- numberguy12
- Registered User
- Join Date: Jan 2017
- Posts: 6,938
- Rep Power: 51915
-
-
Originally Posted By anonkunbrah⏩
Squaring both sides can change the graph of an equation (better to use equation here rather than the word function).so it's because the square root is out of the picture that the restriction lets up and thus will allow for negative values of x? i was baffled since for everything else, it gets the same graph between function rearrangement (y=x-2 is same as x=y+2)
For example, y=x is a basic line with slope 1, going through the origin.
However, y^2= x^2 is a double line, one with slope 1, the other with slope -1, intersecting at the origin. Making an “x”.
You are getting more possibilities with y^2 = x^2. Such as (-3,3), which fits this. The graph of an equation is all the points which satisfy the equation.
Notice that there are ramifications for squaring both sides of an equation: sure (-6)^2 = (6)^2, but this doesn’t mean -6=6.
∫∫ Mathematics crew ∑∑
♫1:2:3:4 Pythagoras crew ♫ ♫ 🧮
Nullius in verba
10-16-2020, 11:41 AM
-
#14
- grey27
- Registered User
-
- grey27
- Registered User
- Join Date: May 2020
- Age: 56
- Posts: 1,675
- Rep Power: 2668
-
-
To generalize the above, if you apply any many-to-one operation to both sides of an equation, then the resulting equation can possibly have solutions that are not solutions to the original equation.
For example
sin(x)=sin(0)
has many more solutions than just x=0.
For example
sin(x)=sin(0)
has many more solutions than just x=0.
10-16-2020, 11:46 AM
-
#15
Originally Posted By AlexSays⏩
could not have explained it betterIn the first:
If you discount imaginary numbers (j), the values of X can only be positive and the value of y cannot exceed 2 or be below -2 or else you end up with the squareroot of a negative number (and hence an imaginary number which can't be plotted here)
In the second:
both x and y can be positive or negative as you are squaring both but again neither can exceed +-2
If you discount imaginary numbers (j), the values of X can only be positive and the value of y cannot exceed 2 or be below -2 or else you end up with the squareroot of a negative number (and hence an imaginary number which can't be plotted here)
In the second:
both x and y can be positive or negative as you are squaring both but again neither can exceed +-2
SuperHercules crew
cancer survivor crew
Dyslexic crew
Friend of Mr.Wilson crew
Ugly and old cell crew
Cat crew
Insomniac crew
10-16-2020, 01:48 PM
-
#16
- numberguy12
- Registered User
-
- numberguy12
- Registered User
- Join Date: Jan 2017
- Posts: 6,938
- Rep Power: 51915
-
-
Originally Posted By snailsrus⏩
Actually this is not really the explanation to OPs question. As mentioned above, x is nonnegative in the above since the first equation is the principal root. Not because of imaginary numbers.could not have explained it better
Also see the above post about applying certain operations to both sides of an equation.
∫∫ Mathematics crew ∑∑
♫1:2:3:4 Pythagoras crew ♫ ♫ 🧮
Nullius in verba
Bookmarks
-
- Digg
-
- del.icio.us
-

- StumbleUpon
-
-
Posting Permissions
- You may not post new threads
- You may not post replies
- You may not post attachments
- You may not edit your posts