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.

- JozelynW

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- JozelynW

How to make a extraneous solution?

- JozelynW

- JozelynW

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## More answers

- JozelynW

do you know how to write a extraneous radical equation

- xapproachesinfinity

extraneous sol is solution that is not in the domain of the equation you started with

- JozelynW

huh

- JozelynW

well, then how do I create a radical extraneous equation?

- 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

it is easier to create an equation with correct solution (not extraneous )

- 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

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

it wants me to use this equation a√x+b+c=d

- xapproachesinfinity

you just said radical equation you did specify any details

- xapproachesinfinity

well at any rate check @welshfella work

- JozelynW

I know because then you would just be giving me answers and then people would be thinking that im cheating

- welshfella

a negative number times a radical will give an extraneous solution

- JozelynW

so i just wanted to know how to create a radical extraneous equation

- JozelynW

can you give me an example

- welshfella

if you plug in 57/2 into the above it won't give you 24

- JozelynW

but it's suppose to

- JozelynW

28.5

- 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

so 28.5 is not 57/2

- welshfella

-3 * sqrt(7 + 28.5) = -17.87

- JozelynW

what do you mean sqrt ? sqrt of 3

- welshfella

do you see that this is an extraneous solution?

- welshfella

square root of (7 + 28.5)

- JozelynW

i do it equals -17.87

- JozelynW

wait did something wrong

- JozelynW

sqrt of 35.5 =5.95 and 5.95 times -3 equals -17.87

- welshfella

yes that is correct
so 28.5 is not a solution its extraneous

- 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

here is a link which deals with the subject
http://www.purplemath.com/modules/solverad2.htm

- JozelynW

what are the options

- welshfella

what do you mean?

- JozelynW

you said " which will fit the original equation"

- 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

sqrt(4*12 + 1) = sqrt 49 = 7

- welshfella

so x = 12 is a valid solution

- JozelynW

so the rule to this is.......

- 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

sorry but i have to go now

- JozelynW

thank you so much for you help @welshfella

- welshfella

yw

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