Question

Given the equation Square root of 2x plus 1 = 3, solve for x and identify if it is an extraneous solution.

^ I really don't understand this topic, whatsoever. Could someone help?

Answer 1

\[\sqrt{2x+1}=3\] right?

Answer 2

get rid of the radical by squaring both sides
you get
\[2x+2=3^2\] i.e. \[2x+1=9\] and solve that one for \(x\) in two steps

Answer 3

typo there, sorry, i meant you get \(2x+1=3^2\) i.e. \(2x+1=9\)

Answer 4

thank you, so the answer is 4, but what exactly does extraneous mean?

two of the answer choices are 4, but one says the solution is extraneous and one says that it isn't.

Answer 5

the answer is 4, and there are no extraneous solutions, that is the one and only answer

Answer 6

i will give you an example of an extraneous solution if you like
when you square both sides, you could introduce a solution to the squared equation that was not a solution to the original equation
that did not happen in this case

Answer 7

okay thanks; so if it were extraneous that's basically saying that there's more than possible one answer for x?

Answer 8

if you plug back x=4 in your original equation, you will indeed get left side = right side.
so there's no extraneous solution.

if you got some value and after plugging it back into original equation, you do not get left side = right side
then that solution is extraneous

Answer 9

you get an extraneous solution sometimes when you square both sides and end up with a quadratic equation that has two solutions, one of them could be wrong

Answer 10

here is a really really simple example
the solution to \(x=-3\) is \(-3\) but if you square both sides you get \(x^2=9\) which, when solving, gives two solutions \(\{-3,3\}\)
but only \(-3\) is a solution to the original equation

Answer 11

i THINK i understand where you getting it..
okay; i have two of those kinds of questions left.

if i give you what i think is the answer, can you like, tell me if i got it right? lol .-.

Answer 12

i can give you another example of one with an extraneous solution if you like
\[\sqrt{x-1}=2x-3\]

Answer 13

ok go ahead
btw if you have answer choices, the easiest thing to do is check the numbers and see which ones work

Answer 14

Solve for x, given the equation Square root of x minus 5 + 7 = 11.

x = 21, solution is extraneous

x = 21, solution is not extraneous

x = 81, solution is extraneous

x = 81, solution is not extraneous

I chose the first one? I got x equals 21, but when I plugged it back into square root of x minus 5 + 7 = 11 it was wrong? sigh .-,

Answer 15

x=21 is correct
when you plugged it back in, how did you get it wrong, try again ?

Answer 16

what is 21- 5 ?

Answer 17

16

but plus 7 is 23?

Answer 18

but there's a square root too!
\(\sqrt{x-5} = \sqrt{21-5}=\sqrt 16 = 4\)

Answer 19

and whats 4+7 ? ;)

Answer 20

oh! lol okay, so then that's not an extraneous solution?

Answer 21

yes! correct :)

Answer 22

okaayy, my last question?

Given the equation −4 Square root of x minus 3 = 12, solve for x and identify if it is an extraneous solution.

x = 0, solution is not extraneous

x = 0, solution is extraneous

x = 12, solution is not extraneous

x = 12, solution is extraneous

not gonna lie I don't know where to start for this one since I don't see an example like this in my lesson.

Answer 23

you can make me work it, just kinda like give me a start

Answer 24

start by dividing both sides by -4

Answer 25

then, you can square both sides

Answer 26

the -4 turns into a 1, the 12 turns into -3

Answer 27

correct

Answer 28

now square both sides

Answer 29

the 1 is gonna stay a 1 though right?

Answer 30

-3 will turn into 9

Answer 31

yess

Answer 32

x = 12

Answer 33

now check whether it is extraneous ?

Answer 34

it is extraneous, right?

Answer 35

absolutely!
now you are good at these ;)

Answer 36

ayeeeeeee! omg thank you so much for the help! lol

including @satellite73

Answer 37

glad i could help :)
welcome ^_^