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anonymous
 5 years ago
Factoring a square root...
sqr(x2)sqr(2x+6)1
I need to turn this into a quadratic equation.
anonymous
 5 years ago
Factoring a square root... sqr(x2)sqr(2x+6)1 I need to turn this into a quadratic equation.

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{x2}\sqrt{2x+6}1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Square everything and solve normally.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ACK! That was so simple... I don't know why I didn't remember that!!!!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now would the negative signs turn to positives?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I mean the central negative sign and the sign in front of 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Nope. All of the quantities are squared. Like, each of the roots and it's (1^2) for the constant.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Okay great, that helped out so much!

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So just to be sure, x=1?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Another 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
 5 years ago
Best ResponseYou've already chosen the best response.0No, it doesn't equal 1, I missed an addition step.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The result would be something like x+3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You get x+3 because if you square everything, you get x2x+61, so combine like terms.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It 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
 5 years ago
Best ResponseYou've already chosen the best response.0That leaves me with: 1\[1\sqrt{12}1=0\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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
 5 years ago
Best ResponseYou've already chosen the best response.0In the original problem, the 1 was =1 and I moved it to the left side, I just didn't include that step.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What was the left side originally?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\sqrt{x2}\sqrt{2x+6}=1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Either way, if I square the 1 on either side, I still end up with 1.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0No, 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
 5 years ago
Best ResponseYou've already chosen the best response.0The last equation I posted was the full equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Then your answer will be an expression, and you cannot move it to the other side.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It says I have to solve for x though, I have to create 0 to solve for x.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh so you mean it was given to you already set equal to one? In the problem they gave you?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, 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
 5 years ago
Best ResponseYou've already chosen the best response.0Lol 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: \[\sqrt{x2}  \sqrt{x+6} = 1\] Which is the same as: \[(x2)^{1/2}  (x+6)^{1/2} = 1\] In order to solve, we must get rid of the root, and we do this by squaring the entire equation. \[(x2)^{(1/2)(2)}  (x+6)^{(1/2)(2)} = 1^{2}. = x  2  x + 6 = 1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Oh! I tried to do that at first, but I didn't think to square those answers.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yeah, just do all that, and then solve for x like a normal linear equation.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Thank you, ahhhh, I was so close :P

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Haha no worries. And I just realized I forgot the 2x so just pretend it's there lol
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