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anonymous

  • one year ago

Solve 8 + √2x less than or equal to 5?

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  1. zepdrix
    • one year ago
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    Hey Alexis :) Is the x under the root like this? \(\large\rm 8+\sqrt{2x}\le 5\)

  2. anonymous
    • one year ago
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    Yes :)

  3. zepdrix
    • one year ago
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    We 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?

  4. anonymous
    • one year ago
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    We would subtract 8

  5. anonymous
    • one year ago
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    From both sides

  6. zepdrix
    • one year ago
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    Mmm k good. That's a good start.\[\large\rm 8+\sqrt{2x}-8\le 5-8\]\[\large\rm \sqrt{2x}\le -3\]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

  7. anonymous
    • one year ago
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    We would square the number by itself to isolate it out?

  8. zepdrix
    • one year ago
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    Good, 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`.

  9. zepdrix
    • one year ago
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    \[\large\rm \left(\sqrt{2x}\right)^2\le \left(-3\right)^2\]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.\[\large\rm 2x\le 9\]

  10. zepdrix
    • one year ago
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    Last step?? :)

  11. anonymous
    • one year ago
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    That is where I am stuck...

  12. zepdrix
    • one year ago
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    Even though you don't see a symbol, the 2 is `multiplying` the x. You need an operation that will undo multiplication.

  13. anonymous
    • one year ago
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    Divide both sides by 2?

  14. zepdrix
    • one year ago
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    Good! :)\[\large\rm \frac{\cancel2x}{\cancel2}\le \frac{9}{2}\]Giving us our final result of\[\large\rm x\le \frac{9}{2}\]

  15. zepdrix
    • one year ago
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    Yayyy team \c:/

  16. anonymous
    • one year ago
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    Omg! thank you so much!!!! I really appreciate your effort and time!

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