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anonymous
 one year ago
Solve 8 + √2x less than or equal to 5?
anonymous
 one year ago
Solve 8 + √2x less than or equal to 5?

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zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hey Alexis :) Is the x under the root like this? \(\large\rm 8+\sqrt{2x}\le 5\)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1We have a bunch of `operations` being applied to x. We need to undo all of them in order to isolate the x. We have some addition, some multiplication, some square root. We'll undo all of that by starting with the most basic. We have an 8 being added, how do we undo that? :O Reverse of addition?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Mmm k good. That's a good start.\[\large\rm 8+\sqrt{2x}8\le 58\]\[\large\rm \sqrt{2x}\le 3\]Hmm the multiplication is `inside` of the root, so we can't get to that operation quite yet. How do we deal with the square root? What is the opposite of square rooting? :O

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We would square the number by itself to isolate it out?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Good, squaring is the inverse of square rooting. Think back to what you did to undo addition. You applied subtraction `to both sides`. We'll simply do that with this step as well. We'll `square both sides`.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \left(\sqrt{2x}\right)^2\le \left(3\right)^2\]So on the left side, the square root and the square `undo` one another, In the same way that +8 and 8 undo one another. On the right side, squaring 3 gives us (3)(3)=9.\[\large\rm 2x\le 9\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is where I am stuck...

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Even though you don't see a symbol, the 2 is `multiplying` the x. You need an operation that will undo multiplication.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Divide both sides by 2?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Good! :)\[\large\rm \frac{\cancel2x}{\cancel2}\le \frac{9}{2}\]Giving us our final result of\[\large\rm x\le \frac{9}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Omg! thank you so much!!!! I really appreciate your effort and time!
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