Question

PLEASE HELP --> WILL GIVE MEDAL --> MUST SHOW WORK

Task 2

Part 1:
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.
a√x + b + c=d
Use a constant in place of each variable a, b, c, and d. You can use positive and negative constants in your equation.
__________________________________________________________
Part 2:
Show your work in solving the equation. Include the work to check your solution and show that your solution is extraneous.
__________________________________________________________
Part 3:
Explain why the first equation

Answer 1

PLEASE HURRY!!! I REALLY NEED HELP!!!

Answer 2

PLS GUIDE ME STEP BY STEP. I WILL DO WORK!

Answer 3

A.)
a√x+b+c=d
has solution:
2√x+5+7=4
has no extraneous solution.
2√x+5+7=11
(b/c 2√x=-1, √x=-.5 x=.25. if you plug it in, you will get a decimal for 2√x)

Answer 4

Whoops I did it for you without steps :(

Answer 5

Can you show steps?

Answer 6

@xo.A

Answer 7

you got that or no?

Answer 8

no :(

Answer 9

k relax, takes a minute, no more

Answer 10

ready?

Answer 11

yes, please! :)

Answer 12

ok your equation is supposed to look like \[a\sqrt{x+b}+c=d\] right ?

Answer 13

yes

Answer 14

and you want two equations, one which has a solution and one which does not (which they call "extraneous")

Answer 15

*thumbs up* yes

Answer 16

are you still there??? @satellite73

Answer 17

first we do one that does have a solution, just make up some numbers for \(a,b,c\) and \(d\) for example \[2\sqrt{x+5}+4=10\]

Answer 18

got hung up,yeah i am here

Answer 19

*nods head*

Answer 20

how do i know it will have a solution?
answer, because the first step is to subtract \(4\) from both sides, and \(10-4=6\) is positive

Answer 21

ok

Answer 22

lets solve it \[2\sqrt{x+5}+4=10\] subtract 4 get \[2\sqrt{x+5}=6\]
divide by 2 get \[\sqrt{x+5}=3\] square both sides get \[x+5=9\]

Answer 23

and last step it so subtract 5 to find \(x=4\)

Answer 24

that is a solution, one that works, and you can check by substituting in back in to the original equation
now want to do one that has only an "extraneous" solution?

Answer 25

I understand the beginning so far, thx!!! But I'm not so sure about this part, it sounds complicated

Answer 26

which part ?

Answer 27

the part with the ''extraneous'' solution, the word is intimidating! :) I'm in a program for 7 graders to do high school math, so ive been a little behind the rest of my class

Answer 28

ok so lets demystify i t

Answer 29

it is possible to solve \[\sqrt{x}=-1\]? the answer is "no"
why not? because \(\sqrt{x}\) is always greater than or equal to zero, so it cannot be negative
not minus 1, minus 2, minus anything

Answer 30

oh, so does extraneous mean that its impossible, kind of like a null set?

Answer 31

you can attempt to solve it, say square both sides and get \[x=(-1)^2\\
x=1\] but that is not a solution

Answer 32

oh

Answer 33

it is math teacherese for "you got an answer, but it don't work"

Answer 34

so now we can come up with an example easy
just make sure at the first step you will get a negative number on the right of the equal sing

Answer 35

sign

Answer 36

ok

Answer 37

make one up, i will go get a drink and come back in two seconds, see if what you had will give an "extraneous" solution

Answer 38

\[2\sqrt{x + 4} -10= 7\]
Does this make sense

Answer 39

I meant plus 10

Answer 40

yes with plus!

Answer 41

\[2\sqrt{x + 4} +10= 7\]

Answer 42

yay!!! Thx!!! :)

Answer 43

first step would be subract \(10\) and get \[2\sqrt{x+4}=-3\]

Answer 44

which is not possible
however let me give you a word of advice

Answer 45

ok

Answer 46

part of the problem is actually to "solve" it even though there solution will be extraneous right ?

Answer 47

yes

Answer 48

so i would start with something like \[2\sqrt{x + 4} +10= 6\] instead of \[2\sqrt{x + 4} +10= 7\] because that way, when you subtrat, you will get an even number

Answer 49

ok, thx. ill keep that in mind

Answer 50

then when you divide by 2 as the third step, you won't have an annoying fraction to work with

Answer 51

second step actually

Answer 52

you happier now?

Answer 53

yes!!! :) Ur a great teacher!!!

Answer 54

(blush)
happy to help
now i am going to go watch Ma and Pa Kettle on TCM
ttytl

Answer 55

bye, Thx again!!! :)

Answer 56

yw( again)