Chrisg094
  • Chrisg094
HELP ME!! with math
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Chrisg094
  • Chrisg094
\[3+\sqrt{2x-3}=x\]
Chrisg094
  • Chrisg094
solve the radical equation.
Chrisg094
  • Chrisg094
Then I need help with another one too.

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Chrisg094
  • Chrisg094
It says, find the domain of the square root function. Write the answer in interval notation \[f(x)=\sqrt{12-4x}\]
Chrisg094
  • Chrisg094
@DariusX can you help me with this problem
Seeing.In.Pixels
  • Seeing.In.Pixels
I don't know if you want it, but I'd be glad to offer my help!
Chrisg094
  • Chrisg094
Give it a shot.
Seeing.In.Pixels
  • Seeing.In.Pixels
Okay, let's start with the first one.
Seeing.In.Pixels
  • Seeing.In.Pixels
What are you trying to do with the equation? Solve for x?
anonymous
  • anonymous
First subtract 3 from both sides, then square it.
Chrisg094
  • Chrisg094
I think it just says solve the radical equation. It doesn't specify much
Seeing.In.Pixels
  • Seeing.In.Pixels
no that will still leave you with two unlike x terms
Chrisg094
  • Chrisg094
@reaganchoi that is smart, total missed that.
Seeing.In.Pixels
  • Seeing.In.Pixels
closet you can get with that is \[6=x ^{2}-x\]
Chrisg094
  • Chrisg094
Oh, wait IDK. I truly need help.
Adrianna.Gongora
  • Adrianna.Gongora
Isn't it 6=x
Seeing.In.Pixels
  • Seeing.In.Pixels
I have a nerd for a dad. His job is math. If you wait just one sec i'll be right back with his answer
Chrisg094
  • Chrisg094
That sounds GREAT!
Adrianna.Gongora
  • Adrianna.Gongora
\[x ^{2}-x\] is equal to x which ends up being x=6
Seeing.In.Pixels
  • Seeing.In.Pixels
Okay, based on this equation, you must be learning quadratics which I hadn't thought o until my dad and I conferred. Anyway, \[ax ^{2}+bx+c\] sounds like your formula for your solution
Chrisg094
  • Chrisg094
That sound about right, I do remember that formula from my last class.
Adrianna.Gongora
  • Adrianna.Gongora
I'm so lost now I thought I had it right... .___.
Chrisg094
  • Chrisg094
I know @Adrianna.Gongora I was doing it the same exact way you was but its different.
Adrianna.Gongora
  • Adrianna.Gongora
I think I did it wrong cause there was no parenthesis around the x part
Chrisg094
  • Chrisg094
its the way @Seeing.In.Pixels is saying
Chrisg094
  • Chrisg094
That is the way its given to me
Seeing.In.Pixels
  • Seeing.In.Pixels
Glad to help. Shall we move on to the second problem, or do you understand?
Chrisg094
  • Chrisg094
No wait, I don't know how to get the problem done. Is it the way @Adrianna.Gongora was saying? or no. I do remember the formula but for some reason I so lost with this problem.
Seeing.In.Pixels
  • Seeing.In.Pixels
Alrighty then. Let's try this again.\[3+\sqrt{2x-3}=x\] We know that we want our answer in the form I mentioned before, right? So let's try to do that. the x's need to be together, so start by squaring both sides to make that possible.
Chrisg094
  • Chrisg094
So the 3 will need to be subtracted to the other side right.
Chrisg094
  • Chrisg094
|dw:1449796816978:dw|
Seeing.In.Pixels
  • Seeing.In.Pixels
After it's been squared, the 3 will be a 9, so you would be subtracting a 9
Chrisg094
  • Chrisg094
|dw:1449796937938:dw|
Chrisg094
  • Chrisg094
oh no wait I just got what you said, I have to square the ( then I would get \[3\sqrt{x}\]
Chrisg094
  • Chrisg094
is that what your saying?
Seeing.In.Pixels
  • Seeing.In.Pixels
no, not square rooting. we're squaring the numbers
Chrisg094
  • Chrisg094
could you just show me what your saying, I think that will help clarify a lot.
Seeing.In.Pixels
  • Seeing.In.Pixels
\[3+\sqrt{2x-3}=x\] Square both sides \[9+2x-3=x ^{2}\] Subtract 9 from both sides \[2x-3=x ^{2}-9\] Add 3 to both sides \[2x=x ^{2}-6\] Subtract 2x from both sides \[0=x ^{2}-2x-6\] Assuming your teacher wants it in the zero form
Seeing.In.Pixels
  • Seeing.In.Pixels
I hope this is what you were looking for. I need to go work on biology, but feel free to message me anytime if you ever need anything
Chrisg094
  • Chrisg094
Thank you so much you were a huge help @seeing.in.pixels I will get back to you
Chrisg094
  • Chrisg094
@Seeing.In.Pixels what about: find the domain of the square root function. write in interval notation.
Chrisg094
  • Chrisg094
\[f(x)=\sqrt{12-4x}\]
Seeing.In.Pixels
  • Seeing.In.Pixels
alright i'm going to look at the problem for a little bit and i'll get back to you in a minute
Chrisg094
  • Chrisg094
Thank you so much, I really do appreciate your help.
Seeing.In.Pixels
  • Seeing.In.Pixels
it's no trouble
Seeing.In.Pixels
  • Seeing.In.Pixels
okay the thing is, you can simplify it, but you can't use interval notation with an equal sign. \[f(x)=4\sqrt{3-x}\]
Chrisg094
  • Chrisg094
so what ever x will equal to say 3 it will look like (-infinity, 3)
Chrisg094
  • Chrisg094
@Seeing.In.Pixels \[f(x)=4\sqrt{3-x}\] is it the final answer
Seeing.In.Pixels
  • Seeing.In.Pixels
x is a mystery value. Interval notation is used when you have a minimum or maximum value for x. if you can pinpoint it, I don't think you need it. I could be wrong, though, so check with your teacher to be sure

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