Question

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

Answer 1

Hey Alexis :)
Is the x under the root like this?

\(\large\rm 8+\sqrt{2x}\le 5\)

Answer 2

Yes :)

Answer 3

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?

Answer 4

We would subtract 8

Answer 5

From both sides

Answer 6

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

Answer 7

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

Answer 8

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`.

Answer 9

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

Answer 10

Last step?? :)

Answer 11

That is where I am stuck...

Answer 12

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

Answer 13

Divide both sides by 2?

Answer 14

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

Answer 15

Yayyy team \c:/

Answer 16

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