Question

6the square root of v. − the square root of the quantity 4 times v squared. − the square root of the quantity 36 times v. + the square root of v squared.?

Answer 1

\[6\sqrt{v}-\sqrt{4v^2}-\sqrt{36v}+\sqrt{v^2}\]

Answer 2

I need to simplify this

Answer 3

@phi

Answer 4

this is similar to your other problems.

have you figured out how to simplify
the 2nd term?
\[ \sqrt{4v^2} \]

Answer 5

Expand it and find pairs?

Answer 6

yes, can you do that ?

Answer 7

\[\sqrt{2\times2 \times v \times v}\]

Answer 8

Pair of twos and pair of v's, then put one on each side?

Answer 9

yes, you "pull out a pair" (leaving a 1 in their place)

Answer 10

\[2v \sqrt{2v+1}?\]

Answer 11

\[ \sqrt{2\cdot 2 \cdot v \cdot v} = 2\cdot v\sqrt{1\cdot 1}= 2v\sqrt{1}= 2v
\]
normally we don't bother with the 1, but it is nice to know it is there.

pay attention...
you see 2*2 inside the parens. you take them out, and put one 2 outside
2 sqrt(v*v)
you see a pair of v's. you pull them out, and put *one* v outside
2*v*sqrt(1)
and of course the square root of 1 is 1, and 2*v*1 is just 2*v or
2v

Answer 12

you now have
\[ 6\sqrt{v}-\sqrt{4v^2}-\sqrt{36v}+\sqrt{v^2} \\6\sqrt{v}-2v-\sqrt{36v}+\sqrt{v^2} \]

can you simplify the 3rd term ?

Answer 13

Put it to the lowest term?

Answer 14

Expand it and find pairs

Answer 15

It's jst 6 and 6

Answer 16

it is
\[ \sqrt{36v}= \sqrt{6\cdot6\cdot v} \]
now pull out any pairs

Answer 17

\[6\sqrt{v}\]

Answer 18

so we now have
\[ 6\sqrt{v}-\sqrt{4v^2}-\sqrt{36v}+\sqrt{v^2} \\6\sqrt{v}-2v-\sqrt{36v}+\sqrt{v^2} \\6\sqrt{v}-2v-6\sqrt{v}+\sqrt{v^2}\]

now the last term

Answer 19

\[v \sqrt{v}\]

Answer 20

?

Answer 21

just \[\sqrt{v} ?\]

Answer 22

you pull out a pair... if there is nothing left, put in a 1 in the place of the pair.
can you do that ?

Answer 23

\[v \sqrt{1} ??\]

Answer 24

yes or just v

you now have
\[ 6\sqrt{v}-2v-6\sqrt{v}+v\]

Answer 25

now combine *like terms*

first, re-arrange to
\[ 6\sqrt{v}-6\sqrt{v}+v-2v \]

can you simplify this ?

Answer 26

factor out the v's?

Answer 27

@phi

Answer 28

what do you get ?

Answer 29

I'm a little confused. Put em all to the side?

Answer 30

\[-12\sqrt{v^2} - 2v\]

Answer 31

The first thing is understand what is a *like term*

\[ 6\sqrt{v}-6\sqrt{v}+v-2v \]

like terms have the same letter to the same power.
that means v and -2v are like terms: both have v to the first power
6 sqrt(v) and -6 sqrt(v) are like terms

Answer 32

can you combine the like terms
6 sqrt(v) -6 sqrt(v)

remember this says you have 6 sqrt(v) take away 6 sqrt(v)

Answer 33

6sqrt v

Answer 34

6 sqrt(v) take away 6 sqrt(v) is ?

Answer 35

6 cows take away 6 cows ?

6 apples take away 6 apples ?

6 strawberries minus 6 strawberries ?

\[ 6 \sqrt{v} - 6 \sqrt{v} = ?\]

Answer 36

Oh, sorry. >.< Not thinking. It's 0

Answer 37

what about the last two terms
v -2v
(notice these are *like terms* because they have the same letter )

Answer 38

2

Answer 39

the v's don't disappear
v - 2v =

1 v - 2 v = ?

Answer 40

1 v take away 2 v's give how many v's ?

Answer 41

-v

Answer 42

so what is the final answer ?

Answer 43

-v?

Answer 44

yes

Answer 45

\[ 6\sqrt{v}-\sqrt{4v^2}-\sqrt{36v}+\sqrt{v^2} = -v\]