Can someone help me solve for x?

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Can someone help me solve for x?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[2\sqrt{x-5}=2\]
I know my first step is to subtract 2 from both sides of the equation which gives me... \[\sqrt{x-5}=0\] Then I subtract 5 from both sides and I'm left with... \[\sqrt{x}=5\] Is there another step after that?
Wooooops! :O

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If the 2 was being `added` to the root, then yes, you would `subtract it from both sides to undo the addition.
But in this case, it looks like the 2 is `multiplying` the root, yes?
So how do we undo multiplication? :)
We divide :D
Good, yes :) Do that. It should give you something a little different than 0 on the right side.
Okay, so now I got 1 instead of 0. @zepdrix
\[\large\rm \sqrt{x-5}=1\]Cool.
Then I would add 5 and get \[\sqrt{x}=6\] @zepdrix
Wooops! The subtraction is `inside` of the root. We can't do that operation yet. We need to deal with the root first.
What's the inverse of `square root`? How do we undo that? :o
Okay so square both sides of the equation? It would still be 1 correct?
\[\large\rm \left(\sqrt{x-5}\right)^2=1^2\]On the left, the operations undo one another, good. On the right, 1^2 = 1? Sounds good!\[\large\rm x-5=1\]
Okay, then I would add 5 to both sides of the equation correct? @zepdrix
yes
Thank you :) @satellite73
yay good job \c:/

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