Find the exact value of the radical expression in simplest form.

Question

anonymous · Answer

$$\sqrt{16y^3} - \sqrt{y^3}$$

anonymous · Answer

I need to know all steps, so I can try to fully understand this.

anonymous · Answer

@johnweldon1993

johnweldon1993 · Answer

Well we can start...by breaking up the first radical...

$$\large \sqrt{16y^3} = \sqrt{16}\sqrt{y^3}$$

We know what the square root of 16 is right? ...

anonymous · Answer

Square root of 16 is 4.

johnweldon1993 · Answer

That is correct...so now we have

$$\large 4\sqrt{y^3} - \sqrt{y^3}$$

right?

johnweldon1993 · Answer

Just to make sure we keep everything organized and together...

$$\large \sqrt{16y^3} - \sqrt{y^3}$$

This became

$$\large \sqrt{16}\sqrt{y^3} - \sqrt{y^3}$$

Which we could then simplify from knowing √16 = 4...

$$\large 4\sqrt{y^3} - \sqrt{y^3}$$

With me?

anonymous · Answer

The second equation, I thought it was different? $$\sqrt{16} \sqrt{y^3}$$ <--- I thought it was this?

johnweldon1993 · Answer

When we wrote that...it was only to simplify the first radical...

when we started with the equation

$$\large \sqrt{16y^3} - \sqrt{y^3}$$

what I did first...was only focus on the first radical...

$$\large \sqrt{16y^3}$$

This is where I got what you wrote....this was simplified as

$$\large \sqrt{16}\sqrt{y^3}$$

But remember...after we found that that was equal to:

$$\large 4\sqrt{y^3}$$


We had to remember that originally we still had that last radical we ignored...so altogether it became

$$\large 4\sqrt{y^3} - \sqrt{y^3}$$

Follow?

anonymous · Answer

Oh okay.

johnweldon1993 · Answer

Yeah...so right now we are at

$$\large 4\sqrt{y^3} - \sqrt{y^3}$$

Right?
Well will you agree that this could also be written as:

$$\large 4\sqrt{y^3} - 1\sqrt{y^3}$$

? Just adding the 1 before that radical...doesn't change anything right?

johnweldon1993 · Answer

Well now...all you do...is (since you have 2 numbers being multiplied to the same square root...we just worry about the numbers in front and NOT change the square root...so

$$\large 4\sqrt{y^3} - 1\sqrt{y^3} = (4 - 1) \sqrt{y^3}$$

So your final answer would be...?

anonymous · Answer

@johnweldon1993 I have the worst laptop in the world. Sorry I had to restart it twice. and. $$3\sqrt{y^3}$$ ?

johnweldon1993 · Answer

lol don't worry about it man...yes that is the correct answer

anonymous · Answer

I have 3 quizs to finish -.- and im writing everything down so i know for the test. and i dont even understand most of them.

anonymous · Answer

In the quiz, that isn't a choice. @johnweldon1993

johnweldon1993 · Answer

What are the choices??

anonymous · Answer

$$3y \sqrt{y}$$
$$3y^2 \sqrt{3}$$
$$4y^2 \sqrt{y}$$
$$4y \sqrt{y}$$

anonymous · Answer

that 3 at the end was supposed to be a y..

johnweldon1993 · Answer

Ahh okay then just 1 more step...didnt think they wanted it THAT simplified...

Okay we have

$$\large 3\sqrt{y^3}$$

Well we can both agree that 

$$\large 3\sqrt{y^3} = 3\sqrt{y^2}\sqrt{y}$$

Well what do we know about the square root of a squared number? It is just the number...so

$$\large 3\sqrt{y^3} = 3\sqrt{y^2}\sqrt{y} = 3y\sqrt{y}$$

And that would be our final answer...

anonymous · Answer

Oh okay. Well now I have to do adding and subtracting ones .-. Thank you.

johnweldon1993 · Answer

Lol no problem man.....and just post a new question if you need help ...If I don't see it I'm sure someone else will help out

anonymous · Answer

I'm writing every question and steps down now to read over them later so I can understand better hopefully. lol.