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

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- anonymous

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

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- zepdrix

Hey Alexis :)
Is the x under the root like this?
\(\large\rm 8+\sqrt{2x}\le 5\)

- anonymous

Yes :)

- zepdrix

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?

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- anonymous

We would subtract 8

- anonymous

From both sides

- zepdrix

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

- anonymous

We would square the number by itself to isolate it out?

- zepdrix

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`.

- zepdrix

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

- zepdrix

Last step?? :)

- anonymous

That is where I am stuck...

- zepdrix

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

- anonymous

Divide both sides by 2?

- zepdrix

Good! :)\[\large\rm \frac{\cancel2x}{\cancel2}\le \frac{9}{2}\]Giving us our final result of\[\large\rm x\le \frac{9}{2}\]

- zepdrix

Yayyy team \c:/

- anonymous

Omg! thank you so much!!!! I really appreciate your effort and time!

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