I Am Awarding Medals! @satellite73 calling @amistre64 @Hero, UNITE! What is the simplified form of the expression 8the square root of v squared. − the square root of v to the fourth power. − the square root of the quantity 9 times v squared. + the square root of the quantity 16 times v to the fourth power.
The answer choices are 15v2 + 11v 11v2 + 15v 5v2 + 3v 3v2 + 5v
@Hero , if you're not too busy do you think you could help?
Or @amistre64 you're good too.
i dont get good grades in my english classes .... even tho i was borned over here
What other language do you speak
i dabble in a few; even samerican sign language, but im not proficient at any of them. English is my strongest tho
Ok, either you'll be fine, I'm sure I'll be able to understand
8the square root of v squared: \(8\sqrt{v^2}\) − the square root of v to the fourth power: \(-(\sqrt v)^4\) − the square root of the quantity 9 times v squared: \(-\sqrt{9v^2}\) + the square root of the quantity 16 times v to the fourth power: \(+\sqrt{16v^4}\)
yes
\[\Large 8v^{2/2}-v^{4/2}-3v^{2/2}+4v^{4/2}\]
That's not the equation
oh, then i guess someone else needs to do it ....
The equation look like this
@Hero any ideas?
Maybe @santiagoto should write what he thinks the equation is.
\[8\sqrt{v^2}-\sqrt{v^4}-\sqrt{9v^2}+\sqrt{16v^4}\]
a few propeties to keep in mind:\[\sqrt{ab}=\sqrt{a}~\sqrt{b}\\~\\\sqrt{k^n}=k^{n/2}\]
ok
\[8v^{2/2}-v^{4/2}-3v^{2/2}+4v^{4/2}\]
@Hero can you show me on vyew, that really helped
Hello?
Try factoring out common factors.
What's common to each term?
\[v ^{2}\]?
\[\sqrt{v^2}\]
Oh
But that simplifies to something
So \[8-\sqrt{v^2}-3+\sqrt{16v^2}\]
What happened to the expression you factored out?
I divided everything by sqrt{v^2}
You have to place it on the outside with parentheses around the expression above.
\[\left(8 - \sqrt{v^2} - 3 \sqrt{16v^2}\right)\sqrt{v^2}\]
What does sqrt(v^2) simplify to?
I was typing that equation in. I thought that Sqrt{v^2} goes in the front
It doesn't matter whether you put in front or back, as long as it's factored out bro. now what does sqrt(v^2) simplify to? C'mon bro.
v
??
Yes. Simplify all of the squares.
Remember the rule \[\sqrt{ab} = \sqrt{a} \dot\ \sqrt{b}\]
Wait, that's kinda confusing
\[\sqrt{16v^2} = \sqrt{16} \dot\ \sqrt{v^2}\] What's confusing about that?
It's the same exact thing I showed you yesterday
Well it was confusing to find the difference of a and b you never showed me that, or did you?
I showed you that rule yesterday.
Oh yeah, lemme get my notes
Got em
Anyway, it simplifies to \[(8 - v - 12v)v\] You should be able to finish simplifying it from there.
16
!!
Is that right? @Hero
I don't know where you get 16 from bro. All you had to do is simplify the expression. It doesn't simplify to a single number.
The expression will include varable v, but it will be in a more simplified form.
I kept going to try other questions and I still don't get it helppppp! @Hero
I thought you were understanding this but I was wrong. If you can't simplify the expression above then you need more help than previously thought.
Not really maybe a quick review anything you got. I understand some but then I get lost and I don't see how some things are done that's when I get tripped up
I don't see why you cannot simplify (8 - v - 12v)v
There's nothing to get tripped up about it. That's basic algebra.
I can do that! Just I was confused before. Remember you're already used to this I'm still just learning it. I'm a rookie and you're a college math guru.
If you can do it, then what is the simplifed version of that expression?
You wrote "16" before which means nothing in terms of simplifying the expression.
8v-v^2-12v^2
And you cannot simplify it further?
8v+12v^4?
No. there's nothing about the expression that suggests to multiply anything.
Isnt that why the v is in the front, for multiplication
7 - 4 = 3 7x - 2x = 5x 5x^2 - 2x^2 = 3x^2
See a pattern there?
When you have like terms you combine them.
No :?
Well, I do...and it's obvious.
But here's the more interesting thing.
7x - 2x = (7 - 2)x = 5x 5x^2 - 2x^2 = (5 - 2)x^2 = 3x^2
Oh
Remember when we were talking about factoring out what's common?
(8 - v - 12v)v has two terms with a common variable. Combine them together first: (8 - 13v)v You can leave it like that or you can multiply the v to get 8v - 13v^2
Either of those last two expressions are acceptable for a final answer.
Oh, I get it thanks again. And you work at a college?
And unfortunately, that isn't one of the answer choices, now what?!
It's not one of the answer choices because I made a mistake.
It's hard not to make a mistake while typing on this computer.
So that's why I got tripped up lol, I knew something was wrong.
My mistakes are usually minor mistakes that could lead to huge mistakes. But your mistakes are just based on not knowing.
Whats up with your computer
There's nothing up with it. I just type too fast sometimes and miss a plus or minus here and there.
Which is not the same as you not understanding how to simplify an expression.
I guess now we can test that out to see if you really understand.
Ok lets go for it. :/
I can show you the mistake I made. I left out a plus: Incorrect: \[\left(8 - \sqrt{v^2} - 3 \sqrt{16v^2}\right)\sqrt{v^2}\] correct: \[\left(8 - \sqrt{v^2} - 3 + \sqrt{16v^2}\right)\sqrt{v^2}\] Now you can try simplifying the correct expression.
(5-sqrt{v^2}+sqrt{16v^2})sqrt{v^2} So far
Bro Did you already forget what sqrt(v^2) reduces to?
And you also seemed to have forgotten what sqrt(ab) equals
v, I know I'm taking it step by step bro
Well, post each step line by line all in the same post. Then submit.
ok (8-v-3+sqrt{16v})sqrt{v} (5+sqrt{17v})sqrt{v}
That's not correct.
\[\sqrt{v^2} \ne \sqrt{v}\]
Also \[\sqrt{16v^2} \ne \sqrt{16v}\]
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