Anyone have a good idea on extraneous and not extraneous equations are? I am working on a project and would like help on figuring out what I am needing to do?

Question

Anyone have a good idea on extraneous and not extraneous equations are?  I am working on a project and would like help on figuring out what I am needing to do?

halfdeafdarling · Answer

Hi, looking for help, can you help?

anonymous · Answer

an extraneous solution is one that you get algebraically, but it is not actually a solution to the equation. What are you trying to do?

halfdeafdarling · Answer

well I have a three part piece I need to be completing.  the first part is this: Create an equation that results in at least one extraneous solution. Work through your equation, justify each step, and explain how the solution is extraneous" I am not sure how to begin with this.

halfdeafdarling · Answer

i figured out thee third piece though well enough so more so two pieces i need help with
--More so i am not sure how to create my own equation

anonymous · Answer

ok. you sometimes get an extraneous solution when solving a radical equation. Something like this, where c and k are constants
$$\sqrt{x \pm c}=x \pm k$$

anonymous · Answer

for example
$$\sqrt{x+2}=x+4$$

anonymous · Answer

do you know how to solve that?

anonymous · Answer

ah, that's a nasty equation. Doesn't actually have any real solutions. Make it
$$\sqrt{x+2}=x-4$$

anonymous · Answer

still here?

halfdeafdarling · Answer

I am I am also going over a khan academy video while reading up on it.  First time viewing it in this situation.  so let me write down your equation, where would I begin on solving it?  Would i square both sides of the equation?

anonymous · Answer

yes

halfdeafdarling · Answer

okay I understand that part of squaring so then we would get.  $$x+4=x^2-16?$$

halfdeafdarling · Answer

there is a part where I believe there should be another number at the end, but that is where i get confused.  Such as the equation on khan academy I was looking at.  $$\sqrt{x}=2x-6$$

halfdeafdarling · Answer

I get through it to where it is x=4x^2-24x+36.  I dont understand on how it is 24 and 36.  once they have 12 do they just multiply that why two then multiply the 6 by itself?  If so why do they do that?

anonymous · Answer

You made a mistake squaring the sides.
$$(\sqrt{x+2})^2=(x-4)^2$$
$$x+2=(x-4)(x-4)$$
$$x+2=x^2-8x+16$$
Then you get everything to one side

anonymous · Answer

in the KA example,
$$(2x-6)^2 = (2x-6)(2x-6)$$
foil to get
$$4x^2-24x+36$$

halfdeafdarling · Answer

I think my mind is making this more difficult than it should be.  I see how to get the x^2 but the -8x is still confusing (im understanding the 16 though because the distributing 4*4=16 .  the -8x where is that?  (sorry if i seem like it is going over my head first time seeing this being done really confused)

halfdeafdarling · Answer

OOOH is it adding like terms?

halfdeafdarling · Answer

cause you are getting -4x on each side which would be getting the -8x ?

anonymous · Answer

It's FOIL/distribution

$(x-4)(x-4)=x^2-4x-4x+16=x^2-8x+16$

anonymous · Answer

exactly

halfdeafdarling · Answer

oh okay wow sorry about that, that went over my head there.

halfdeafdarling · Answer

so then we would have $$x+2=x^2-8x+16$$

halfdeafdarling · Answer

then we would be subtracting two ?

anonymous · Answer

yes. subtract 2 and x. You need one side to be 0 to solve the quadratic equation

halfdeafdarling · Answer

okay got that so then it would be 0=x^2+6

anonymous · Answer

no. not sure what you did. You actually need to combine like terms.
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