Question

How do i simplify radical expressions?

Answer 1

what radical question

Answer 2

the square root of 125n

Answer 3

you do 125/6
then
9*9

Answer 4

\[\large \sqrt{125n}\]

Hint
\[\large 125 = 25 \times 5\]

Answer 5

So we might really have

\[\large \sqrt{25 \times 5 \times n}\]

Answer 6

n=5 you twat

Answer 7

so the answer would be...?

Answer 8

5 babe dont listen to them

Answer 9

so the n doesnt matter? @jeedeee

Answer 10

\[\large \sqrt{25 \times 5 \times n}\]

the square root of 25 is 5 so we have

\[\large 5\sqrt{5n}\]

Answer 11

so thats the answer? hah

Answer 12

the answer is 2.5 n=2.5

Answer 13

yeah @Mychelynn99

Idk where @jeedeee was looking...all you asked was to simplify √125n Idk where the 6 or 9*9 came from?

Answer 14

sooooo the answer is 5n?

Answer 15

The answer would be \[\large 5\sqrt{5n}\]

Answer 16

Okay thanks!

Answer 17

So we have

\[\large \sqrt{512k^2}\]

We know that \[\large 512 = 256 \times 2\]

so we really have \[\large \sqrt{256 \times 2 \times k^2}\]

we know that the square root of 256 = 16 so we have

\[\large 16\sqrt{2 \times k^2}\]
and we also know that \[\large \sqrt{k^2} = k\]

so alltogether we have

\[\large 16k\sqrt{2}\]

Answer 18

@Mychelynn99 :)

Answer 19

so if i have the square root of 80p^3 the answer would be.... umm yeah i still dont fully understand @johnweldon1993

Answer 20

Alright no problem

Alright so notice how first...I take the number (80 here) and break it up

We know:
40 times 2 = 80
20 times 4 = 80
16 times 5 = 80 <--- right here is where we stop

Why? because we know 16 is a perfect square...

Answer 21

Perfect square meaning

1^2 = 1
2^2 = 4
3^2 = 9
4^4 = 16 <--- look at that

Answer 22

So how does that help?

Well...if we know that 80 can be broken up into 16 times 5....we can write it as

\[\large \sqrt{16 \times 5 \times p^3}\]

Answer 23

And as we just found out....the square root of 16...is 4....since 4 squared.. 16

\[\huge \sqrt[2]{4^2} = 4\]

Answer 24

So now what we have....knowing that the square root of 16...is 4....is

\[\large 4\sqrt{5p^3}\]

Answer 25

Making sense so far hun?

Answer 26

kinda...

Answer 27

Which part is confusing? :)

Answer 28

how to know to stop at 16 times 5

Answer 29

Oh okay....

alright firstly....we need to understand perfect squares...

meaning when you square a number....(multiply it by itself) what does it equal?

so for example

when you square 1....(multiply 1 by itself) so \[\large 1 \times 1 = 1\]

so \[\large 1^2 = 1\]

does that make sense?

Answer 30

yes

Answer 31

Great....

so if we continue on that path

2 squared would be

\[\large 2^2 = 2 \times 2 = 4\]
\[\large 3^2 = 3 \times 3 = 9\]
\[\large 4^2 = 4 \times 4 = 16\]

okay?

Answer 32

Got it

Answer 33

Alright...so back to the original problem

lets ignore the p^3 for this part...

imagine we just have

\[\large \sqrt{80}\]

Like we said...we want to figure out if any perfect squares divide evenly into 80

Answer 34

40 times 2?

Answer 35

We do that...because the square root....of a number squared....is the number

for example...

\[\large \sqrt{1^2} = 1\]
\[\large \sqrt{5^2} = 5\] etc...the SQUARE root and the SQUARE cancel out..leaving us with just the number

Answer 36

So yes...lets list what goes into 80 evenly

If we multiply 40 times 2...we get 80.....but is either 40 or 2 a perfect square? Nope...so we keep going...

if we multiply 20 times 4....we get 80....now we see that 4 is indeed a perfect square \(\large 2^2 = 4\) so we CAN do that...but...we see that 20 can still be broken down a bit so we can keep going...

(we CAN stop there though....but we wont :)

Answer 37

So lets continue now...

we know that 16 times 5 = 80 ....and we want to stop here.....because we know that 16 is a perfect square \(\large 4^2 = 16\) and we can see that 5 cannot be broken down anymore...

Answer 38

got it so far

Answer 39

Great....so lets start from the beginning..

\[\large \sqrt{80p^3}\]

Now that we know 80 can be written as 16 times 5...lets write it as so...

\[\large \sqrt{16 \times 5 \times p^3}\]

Answer 40

We just found...that 16...is a perfect square....it is \(\large 4^2 = 16\)

since that is the case...we can write the 4 on the outside

\[\large 4\sqrt{5p^3}\]

still good?

Answer 41

yupp

Answer 42

Great....now we focus on the \(\large p^3\)

We know that \[\large p^3 = p^2 \times p\] right?

Answer 43

right

Answer 44

So lets write that

\[\large 4\sqrt{5 \times p^2 \times p}\]

okay?

Answer 45

Remember how the square root of a squared number...is just the number?

\[\large \sqrt[2]{7^2} = 7\]
and
\[\large \sqrt[2]{9^2} = 9\]

well the same rule stands here...just imagine 'p' was a number...so

\[\large \sqrt[2]{p^2} = p\]

make sense?

Answer 46

clear as day

Answer 47

Great!

So just like when we wrote the 4 on the outside when we simplified \(\large \sqrt[2]{16}\) we can now write 'p' on the outside since we simplified that p^2

so altogether we have

\[\large 4p\sqrt{5p}\]

Answer 48

And that would be our final answer there :)

Answer 49

ugh this is so complicated!

Answer 50

What confused you there hun?

Answer 51

why is there a p by the 5 too?

Answer 52

Well remember we had

\[\large 4\sqrt{5 \times p^3}\]

and we knew that \[\large p^3 = p^2 \times p\]

so when we wrote tht in...we had

\[\large 4\sqrt{5 \times p^2 \times p}\]

Answer 53

So when we simplified \(\large \sqrt{p^2}\) we do indeed get the 'p' that showed up outside the square root...but notice how we never touched that last 'p' that stays inside the square root

\[\large 4p\sqrt{5p}\]

Answer 54

gotcha

Answer 55

sorry imm askng so much

Answer 56

No not at all hun :) I want to make sure you get this :)

Answer 57

So..can you help me on another?

Answer 58

Of course :)

Answer 59

the square root of 45p^2

Answer 60

Alright...so same thing here...

what times what = 45? list anything you can think of...

Answer 61

7*5

Answer 62

but thats not a perfect square

Answer 63

\[\large 7 \times 5 \cancel{=}45\]

Answer 64

LMAO blonde moment 9*5

Answer 65

Haha we all have em ;P

Okay....good :) ....now...does that have a perfect square?

Answer 66

nope

Answer 67

.....sure? ... :)

hint*

\[\large 3^2 = ?\]

Answer 68

ohh 3*3=9

Answer 69

Mmhmm :)

so we know that 9 is a perfect square :)

\[\large \sqrt{45p^2}\]

we found that 45 can be written as 9 times 5 so

\[\large \sqrt{9 \times 5 \times p^2}\]

And we just saw that 9 is \(\large 3^2\) so we write that 3 on the outside like before...

\[\large 3\sqrt{5p^2}\]

making sense?

Answer 70

makes sense. I think im finally catching on. omg but how would i do square root of 28x^3y^3

Answer 71

Focus on 1 part at a time :)

the 28 part...what times what = 28?

Answer 72

7 times 4

Answer 73

so it would be 7*4*x^3*y^3?

Answer 74

RIGHT :)

so we have that

\[\large \sqrt{7 \times 4 \times x^3 \times y^3}\]

wheres the perfect square?

Answer 75

4 2*2

Answer 76

Right! :D good you are getting it :D

so write that 2 on the outside...

\[\large 2\sqrt{7 \times x^3 \times y^3}\]

remember before...how we knew that \(\large p^3 = p^2 \times p\) ?

Answer 77

im kinda lost

Answer 78

Where hun?

Answer 79

the p hah

Answer 80

lol I was just bringing up that last problem...

where we had \[\large p^3\]

and we knew we can write that as \[\large p^2 \times p\]

Answer 81

oh yeah haha

Answer 82

Alright lol

so here we have that same thing...but we have

\[\large x^3\] and \[\large y^3\]

so those can be written as ?

Answer 83

so x^2*x?

Answer 84

Mmhmm...anddddddd (the 'y' one now :)

Answer 85

y^2*y

Answer 86

Great!

Alltogether we have

\[\large 2\sqrt{7 \times x^2 \times x \times y^2 \times y}\]

right?

Answer 87

Right

Answer 88

Now remember what happened last time?

\[\large \sqrt[2]{p^2} = p\]

sooo here when we simplify both the x^2 and the y^2...we get?

Answer 89

umm not sure

Answer 90

well it would be the same thing that happened before....I was bringing that 'p' there again to show what happened...

since we knew that √p^2 became p

we know that √x^2 will be x
and that √y^2 will be y right?

Answer 91

oh psh duh okay so x^2 would be x, right? and same with y

Answer 92

Mmhmm..so lets catch up...we had

\[\large 2\sqrt{7 \times x^2 \times x \times y^2 \times y}\]

and now we know...when we simplify the x^2 and the y^2...we will have an 'x' and a 'y' on the outside again....but remember...we never touch the other x and the other y that are in there...

so alltogether we have

\[\large 2xy \sqrt{7xy}\]

Answer 93

alrighty

Answer 94

But...does that make sense?

Answer 95

yes

Answer 96

Awesome :)

Answer 97

Feel free to message me if you need any more help :)