anonymous
  • anonymous
Factoring a square root... sqr(x-2)-sqr(2x+6)-1 I need to turn this into a quadratic equation.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\sqrt{x-2}-\sqrt{2x+6}-1\]
anonymous
  • anonymous
Square everything and solve normally.
anonymous
  • anonymous
ACK! That was so simple... I don't know why I didn't remember that!!!!

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anonymous
  • anonymous
Now would the negative signs turn to positives?
anonymous
  • anonymous
I mean the central negative sign and the sign in front of 1
anonymous
  • anonymous
Nope. All of the quantities are squared. Like, each of the roots and it's -(1^2) for the constant.
anonymous
  • anonymous
Okay great, that helped out so much!
anonymous
  • anonymous
So just to be sure, x=-1?
anonymous
  • anonymous
Another way to think of it would be to write out the exponents of each quantity, like the root is to the one half, then if you multiply (1/2)(2), you get (2/2)=1 right? So the exponents cancel but the quantity inside stays the same.
anonymous
  • anonymous
No, it doesn't equal -1, I missed an addition step.
anonymous
  • anonymous
The result would be something like -x+3
anonymous
  • anonymous
x=3
anonymous
  • anonymous
You get -x+3 because if you square everything, you get x-2-x+6-1, so combine like terms.
anonymous
  • anonymous
It seems right until I plut it into \[\sqrt{2(3)+6}\] That turns into:\[\sqrt{12}\] But there is no complete square root of 12, so whatever it is -1, it can't be right.
anonymous
  • anonymous
That leaves me with: 1-\[1-\sqrt{12}-1=0\]
anonymous
  • anonymous
No, that's not it. The result is not a number unless you have something to set it equal to. You get an expression as an answer.
anonymous
  • anonymous
In the original problem, the -1 was =1 and I moved it to the left side, I just didn't include that step.
anonymous
  • anonymous
What was the left side originally?
anonymous
  • anonymous
\[\sqrt{x-2}-\sqrt{2x+6}=1\]
anonymous
  • anonymous
Either way, if I square the 1 on either side, I still end up with -1.
anonymous
  • anonymous
No, I mean the full equation. In your question, you gave an expression. An equation must have an equal sign. Was zero on the other side of the equation originally? Or was nothing there?
anonymous
  • anonymous
The last equation I posted was the full equation.
anonymous
  • anonymous
Then your answer will be an expression, and you cannot move it to the other side.
anonymous
  • anonymous
It says I have to solve for x though, I have to create 0 to solve for x.
anonymous
  • anonymous
Oh so you mean it was given to you already set equal to one? In the problem they gave you?
anonymous
  • anonymous
Yes, exactly. I just left out that step in my original post. The intent was to save you some work, but I will know better next time :P
anonymous
  • anonymous
Lol well you have to say that you need to solve for x, otherwise the answer will be an expression. Then here's what you do: \[\sqrt{x-2} - \sqrt{x+6} = 1\] Which is the same as: \[(x-2)^{1/2} - (x+6)^{1/2} = 1\] In order to solve, we must get rid of the root, and we do this by squaring the entire equation. \[(x-2)^{(1/2)(2)} - (x+6)^{(1/2)(2)} = 1^{2}. = x - 2 - x + 6 = 1\]
anonymous
  • anonymous
Oh! I tried to do that at first, but I didn't think to square those answers.
anonymous
  • anonymous
Yeah, just do all that, and then solve for x like a normal linear equation.
anonymous
  • anonymous
Thank you, ahhhh, I was so close :P
anonymous
  • anonymous
Haha no worries. And I just realized I forgot the 2x so just pretend it's there lol

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