What would y be?
^3√(3y+3)=3

- anonymous

What would y be?
^3√(3y+3)=3

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- anonymous

Well the idea is to isolate y right? What would be your first step?

- anonymous

You have a cube root. How would you get rid of that?

- anonymous

For instance, if you have a square root you have to square to get rid of the square root.

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## More answers

- anonymous

@francescazares2012 : You there?

- anonymous

yes

- anonymous

So I would cube both sides?

- anonymous

Yep!

- anonymous

Is my answer 2?

- anonymous

Good job!

- anonymous

It didn't work. :/

- anonymous

the answer is 8! :D

- anonymous

Ohh wait. I am dumb.

- anonymous

3^3 is 27 >.< .

- anonymous

thanks for your help! I was having so much trouble

- anonymous

Yeah it's 8 :P .

- anonymous

could you help me with another?

- anonymous

Sure.

- anonymous

√(x+10)=x+10

- anonymous

Okay, square both sides again :) .

- anonymous

okay so now i have x+10=(x+10)^2

- anonymous

Can you expand the (x+10)^2 ?

- anonymous

yes

- anonymous

|dw:1355462739047:dw|

- anonymous

okay I am a bit confused

- anonymous

With?

- anonymous

what I am supposed to do

- anonymous

Okay so we have:
|dw:1355462917095:dw|

- anonymous

|dw:1355463044826:dw|

- anonymous

okay I sort of see where this is going

- anonymous

You realize we have to foil this correct?

- anonymous

|dw:1355463252789:dw|

- anonymous

okay

- anonymous

What did you get?

- anonymous

-10 and -9?

- anonymous

Very good! :) .

- anonymous

wow!

- anonymous

Algebra isn't that bad once you get the hang of it :) .

- anonymous

I have a algebra final tomorrow. I am pretty nervous.

- anonymous

Haha, you will be fine ^_^ .

- anonymous

I am working on this problem right now and I got 3 but it says it's wrong.
√(6a-2)=16

- anonymous

not 16 I meant -4

- anonymous

so √(6a-2)=-4

- anonymous

So square both sides :) .

- anonymous

Keep the negative inside the brackets when you square, so it would be like (-4)^2 .

- anonymous

but wouldn't that be 16?

- anonymous

Yep :) .

- anonymous

A genral rule is when you square both sides and the negative it attached to a number, You put the negative in brackets and them square it.

- anonymous

Because when you square you are basically going -4 * -4 which his 16 ( negative * negative is +) .

- anonymous

Okay. Well I got the final answer to be 3. But it says that that is wrong. Why is that?

- anonymous

So when you square both sides, what did you get?

- anonymous

It's strange, that's absolutely correct.

- anonymous

That's what I thought. It's online hw. And it tells you when it's right or wrong.

- anonymous

nevermind it was does not exist

- anonymous

probably because it was a -4

- anonymous

No, that's not why.

- anonymous

It depends on the sign of the square root.

- anonymous

If it's a POSITIVE square root then yes, no solutions exist.

- anonymous

But if it's -√(6a-2) then it's 3.

- anonymous

Because when you take the square root of a function it's always in the form of ±√ f(x) . We need to know the proper sign.

- anonymous

That question was badly worded.

- anonymous

You can say that again!

- anonymous

Anything else?

- anonymous

Only if you are up for it!

- anonymous

√(x+6)=√(x)-6

- anonymous

Square both sides :P .

- anonymous

I found the answer

- anonymous

It's a procedure essentially :P .

- anonymous

What did you get?

- anonymous

does not exist

- anonymous

Good job!

- anonymous

√(y+1)=√(y)+1

- anonymous

what about this one?

- anonymous

Square both sides :) .

- anonymous

okay

- anonymous

y+1=y+1?

- anonymous

No....

- anonymous

We have:
|dw:1355464812127:dw|

- anonymous

You square the WHOLE thing :) .

- anonymous

oh okay

- anonymous

so what do i do now?

- anonymous

After squaring both sides?

- anonymous

yes

- anonymous

Get all the y's on one side.

- anonymous

I think I am having trouble squaring them.

- anonymous

|dw:1355465159249:dw|

- anonymous

oh so I foil

- anonymous

You should get the same answer no matter what you use.

- anonymous

oh okay

- anonymous

What did you do?

- anonymous

I am moving the y's

- anonymous

I think I am doing this wrong

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