Question

Please do me a favor... Fan+Medal :)

Answer 1

Create me an extraneous solution with this:
a(√x+b)+c=d

Answer 2

don't know

Answer 3

@hero can you help me?

Answer 4

what about you @shiraz14

Answer 5

@HenrietePurina : An extraneous solution arises when you multiply both sides by a variable. Can you suggest how we may create one in this context?

Answer 6

well, ok and i need it to have no solution... my idea was:
5(√x+2)+9=1

Answer 7

5(√x+2)+9=1
5(√x+b)=-8
???

Answer 8

oops, the b was supposed to be 2

Answer 9

Well, you can do that ... let's see what you get:

5(√x+2)=-8
5(√x+2)(√x-2)=-8(√x-2)
5(x-4)=-8(√x-2)

Answer 10

Now, let u=√x, and sub. u into the equation.

Answer 11

ok, but I want you to know that the equation is actually this\[5\sqrt(x+2)=-8\]

Answer 12

Does that make a difference?

Answer 13

OK, yes, it does make a difference. So your equation is actually 5[√(x+2)]=-8?

Answer 14

Ahhh... ok sorry i should have clarified that earlier... sorry

Answer 15

In that case, you might want to multiply both sides by √(x+2).

Answer 16

and no that is just an equation i thought of myself but really we need a extranous (no solution) with the base of \[(a)\sqrt(x+b)+c=d\]

Answer 17

I don't think I get your question. Are you looking for an extraneous solution to a given equation or are you required to create an equation which has an extraneous solution?

Answer 18

i'm supposed to create one, again im so sorry for being so confusing

Answer 19

OK, & the equation you create will have to be of the form (a)√(x+b)+c=d ?

Answer 20

yup :) and it has to have an extraneous solution

Answer 21

Ok, in that case, a, b & c can take on any non-zero values (I prefer to use positive integers for simplicity) while d=0.

Answer 22

ok :)

Answer 23

An example (using your previous example):
5[√(x+2)]+8=0

Answer 24

To create an extraneous solution for this case, multiply both sides of the equation by √(x+2).

Answer 25

5 + 8 = \[\sqrt{}\]

Answer 26

oops...\[5 + 8 = \sqrt{x+2}\]

Answer 27

Come to think of it, it would be better if you let a=-5 instead. Then, you would get:

-5(x+2)+8√(x+2)=0

Then substitute u=√(x+2) into the above equation to get:
-5u² + 8u = 0
Solving for u gives:
u=1.6 or u=0
When u=1.6, x=0.56
When u=0, x=-2 (this is an extraneous solution)

Answer 28

thanks, wow, it was that easy... hehe thanks bro a gazillion times :)

Answer 29

You're welcome - so your equation would be something like -5[√(x+2)]+8=0.

Answer 30

is that the same as -5√(x+2) + 8 = 0?

Answer 31

\[-5 \sqrt{x+2}+8=0 ?\]

Answer 32

Yes, it is. I'm using the square braces just to avoid confusion between the 5th root and the constant 5 multiplied by the square root (which is what we are referring to here), simply because I'm not using the Equation tool in Open Study.

Answer 33

oops i meant that just copy paste failed :)

Answer 34

ok :)

Answer 35

You're welcome