JozelynW
  • JozelynW
Create two radical equations: one that has an extraneous solution, and one that does not have an extraneous solution. Use the equation below as a model.
Mathematics
schrodinger
  • schrodinger
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JozelynW
  • JozelynW
How to make a extraneous solution?
JozelynW
  • JozelynW
JozelynW
  • JozelynW

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JozelynW
  • JozelynW
do you know how to write a extraneous radical equation
xapproachesinfinity
  • xapproachesinfinity
extraneous sol is solution that is not in the domain of the equation you started with
JozelynW
  • JozelynW
huh
JozelynW
  • JozelynW
well, then how do I create a radical extraneous equation?
xapproachesinfinity
  • xapproachesinfinity
consider the following equation \(\sqrt{\frac{x+1}{x}}=\sqrt{\frac{1}{x}}\) see that x=0 is not in the domain try to solve that equation
xapproachesinfinity
  • xapproachesinfinity
it is easier to create an equation with correct solution (not extraneous )
welshfella
  • welshfella
-3 sqrt(7 + 2x) = 24 sqrt(7 + 2x) = -8 square both sides 7 + 2x = 64 2x = 57 x = 57/2 now plug in x = 57/2 and see if it satisfies the original equation
xapproachesinfinity
  • xapproachesinfinity
consider the following equation \(\sqrt{x+1}=\sqrt{3x-1}\) solution is \(x=1\) if we check it on the equation we it checks out
JozelynW
  • JozelynW
it wants me to use this equation a√x+b+c=d
xapproachesinfinity
  • xapproachesinfinity
you just said radical equation you did specify any details
xapproachesinfinity
  • xapproachesinfinity
well at any rate check @welshfella work
JozelynW
  • JozelynW
I know because then you would just be giving me answers and then people would be thinking that im cheating
welshfella
  • welshfella
a negative number times a radical will give an extraneous solution
JozelynW
  • JozelynW
so i just wanted to know how to create a radical extraneous equation
JozelynW
  • JozelynW
can you give me an example
welshfella
  • welshfella
if you plug in 57/2 into the above it won't give you 24
JozelynW
  • JozelynW
but it's suppose to
JozelynW
  • JozelynW
28.5
welshfella
  • welshfella
no - ones like that will give an extraneous solution When you square you get a positive result so a negative value can get 'lost'.
JozelynW
  • JozelynW
so 28.5 is not 57/2
welshfella
  • welshfella
-3 * sqrt(7 + 28.5) = -17.87
JozelynW
  • JozelynW
what do you mean sqrt ? sqrt of 3
welshfella
  • welshfella
do you see that this is an extraneous solution?
welshfella
  • welshfella
square root of (7 + 28.5)
JozelynW
  • JozelynW
i do it equals -17.87
JozelynW
  • JozelynW
wait did something wrong
JozelynW
  • JozelynW
sqrt of 35.5 =5.95 and 5.95 times -3 equals -17.87
welshfella
  • welshfella
yes that is correct so 28.5 is not a solution its extraneous
welshfella
  • welshfella
now one that does not have an extraneous solution would be sqrt(4x + 1) = 7 so 4x + 1 = 49 4x = 48 x = 12 which will fit the original equation
welshfella
  • welshfella
here is a link which deals with the subject http://www.purplemath.com/modules/solverad2.htm
JozelynW
  • JozelynW
what are the options
welshfella
  • welshfella
what do you mean?
JozelynW
  • JozelynW
you said " which will fit the original equation"
welshfella
  • welshfella
I meant if you replace x by 12 in the original equation then the equation will balance the left side will equal 7
welshfella
  • welshfella
sqrt(4*12 + 1) = sqrt 49 = 7
welshfella
  • welshfella
so x = 12 is a valid solution
JozelynW
  • JozelynW
so the rule to this is.......
welshfella
  • welshfella
a negative value as above will give an extraneous solution. But there' more to it than that that's why i posted the link. Some equations will have one valid and one extraneous solution , for example.
welshfella
  • welshfella
sorry but i have to go now
JozelynW
  • JozelynW
thank you so much for you help @welshfella
welshfella
  • welshfella
yw

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