Question

what is x^2 - 18x + 82 factored?

Answer 1

HI!!

Answer 2

i bet it is really
\[x^2-18x+81\] right? not \(82\)

Answer 3

the thing is i'm doing radicals so I had to move the one to the other side, so now i'm kind of stuck on what to do

Answer 4

if it is what i wrote, then it is a perfect square, namely
\[(x-9)^2\]

Answer 5

oh dear, there is no 'other side" in your question, because you do not have an equal sign there

Answer 6

Okay lets start over, maybe you could help me. Can i send you the question I am doing?

Answer 7

is it
\[x^2-18x+82=0\]?

Answer 8

sure go ahead

Answer 9

No lol here just take a look
When solving a radical equation, John and Tim came to two different conclusions. John found a solution, while Tim's solution did not work in the equation.

Create and justify two situations: one situation where John is correct and a separate situation where Tim is correct.

Answer 10

hmm that is a different question entirely

Answer 11

a radical equation could look like
\[\sqrt{x-1}+3=5\]

Answer 12

you would solve that by writing
\[\sqrt{x-1}+3=5\\
\sqrt{x-1}=2\\
x-1=4\\
x=5\]

Answer 13

Yes exactly, I already had one written, but then I got to x = x^2 - 18x +82 and got totally confused.

Answer 14

you're suppose to square both sides?

Answer 15

okay, sometimes when you solve an equation that has a radical sign in it, by squaring it, you end up introducing what is called an extraneous solution. An extraneous solution is not a mistake, but rather a solution that doesn't work in the original equation. There's a video here that might help you:
https://www.khanacademy.org/math/algebra/exponent-equations/radical_equations/v/extraneous-solutions-to-radical-equations

Answer 16

Okay, i see what you did

Answer 17

yes squared both sides

Answer 18

@Here_to_Help15 I know, I've seen that already, I'm just stuck on finding the right problem that will give me a extraneous and non-extraneous solution

Answer 19

but if you had
\[\sqrt{x-1}+5=3\] then
\[\sqrt{x-1}=-2\] and if you square you get
\[x-1=4\] so
\[x=5\] but \(5\) is not really as solution, you can check it

Answer 20

the first one i wrote has a real solution, the second one has an "extraneous" solution

Answer 21

It has to have both, see this was my equation \[\sqrt{x-1} = x-9\]

Answer 22

but it didn't work out, it got ugly lol

Answer 23

Unusual question!
Let's go with Tim first. Suppose Tim encounters the radical equation
\[\sqrt{x} + 9=0\]
Tim tries to solve this by moving the 9 to the other side of the equation; he subtracts 9 from both sides and obtains
\[\sqrt{x} = - 9\]

To eliminate the radical and to solve for x, he now squares both sides of this equation. Can you help him do that?
What possible solution do the two of you arrive at?
Substitute this possible solution back into the original equation. Is the resulting equation true or false?
How about you try supervising John. Guide him in the right direction to create and solve a radical equation whose solution really is a solution, that is, it satisfies the original equation.

Answer 24

You're copying that right off someone else on here. I've read that one already. Thanks for trying though.

Answer 25

As you know im quoting this since i have very few knowledge on what this question is asking i will provide link
http://openstudy.com/study#/updates/52c5ce14e4b0b729fb8cc9fe

Answer 26

I know, but it didn't help me lol

Answer 27

Yea :) Sorry i should have said where i was getting it from first

Answer 28

No, it's okay . Radicals aren't my strong suit lol i'll figure it out, but thank you for trying!