Question

simplified form sqrt 16c^4 - sqrt c^2 + 3 sqrt c^2 + 9c^2
PLease help

Answer 1

\[\sqrt{16c ^{4}}-\sqrt{c ^{2}}+ 3\sqrt{c ^{2}}+ \sqrt{9c ^{2}}\]

Answer 2

most of the terms are perfect square, we can extract their square roots and make it all in c variable...

Answer 3

can you show me?

Answer 4

... except the 1st term...

Answer 5

\[\sqrt{16c^4}=\sqrt{16}\times\sqrt{c^4}=4\times c^2\]

Answer 6

or simply \(4c^2\)

Answer 7

... we do the same for the rest, you think you can make it?

Answer 8

lets see..
is 4c^2 = 2c?

Answer 9

no, we have already removed the radical sign for the 1st term so it's already done, i'm asking for the other terms (2nd to the 4th)...

Answer 10

the second term would be c..?
and the last term would be 9c..?
and im ify on the third term ...

Answer 11

ok for the 2nd term, the last term? 9 is under the radical sign and a perfect square so what is its square root?

Answer 12

3

Answer 13

so what is the simplified for the last term?

Answer 14

3c

Answer 15

ok you're getting it all right... for the 3rd term, it is c^2 in a radical sign, what is the square root of c^2?

Answer 16

c

Answer 17

ok you're correct again, now write all the simplified terms we obtain based on their arrangement above... see if we can further simplify it...

Answer 18

starting with the 1st term...

Answer 19

\[4c ^{2}-c+c+3c\]

Answer 20

you forgot to include the constant 3 in the 3rd term...

Answer 21

... actually you can simplify the 2nd to the last term since they are all common to c variable...

Answer 22

\[4c ^{2}-3c+c+3c\]

Answer 23

like that ?

Answer 24

it should be \[4c^2-c+3c+3c\]

Answer 25

so would the final answer be \[4c ^{2}+7c\]

Answer 26

no, take note the sign of the 2nd term, it is -c

Answer 27

so \[4c ^{2}+5c\]

Answer 28

yes you're correct... you got it... :)

Answer 29

thanks can you help me with one more ?! sorry im so freaking slow..

Answer 30

ok no problem... :)

Answer 31

\[5\sqrt{13} +6 -7 \sqrt{13}\]

Answer 32

i need to find the exact value

Answer 33

do you think you can simplify further the \(\sqrt{13}~?\)

Answer 34

no

Answer 35

if you will look at the expression, do you think you can combine similar terms and perform their operations?

Answer 36

what are those terms?

Answer 37

\[13\sqrt{13}\]

Answer 38

wait no

Answer 39

\[-2\sqrt{13}\]

Answer 40

remember you have 6 alone...

Answer 41

\[-2\sqrt{13} + \sqrt{6}\]

Answer 42

no it's not, look at the 6, it is under a radical sign?

Answer 43

no

Answer 44

then how would you write it now?

Answer 45

ok it is \[6-2\sqrt{13}.\]