Question

Multiplying radicals
5√6x2x√6
(addition problem)

Answer 1

@jim_thompson5910

Answer 2

Please help

Answer 3

so you want to multiply \(\large 5\sqrt{6}\) with \(\large 2\sqrt{6}\) right?

Answer 4

Correct

Answer 5

ok what you do is multiply the coefficients to get: 5*2 = 10

then you multiply the terms inside the square root: 6*6 = 36

Answer 6

so that will turn

\(\large 5\sqrt{6}*2\sqrt{6}\)

into

\(\large 10\sqrt{36}\)

Answer 7

from here, you take the square root of 36 to get 6

Answer 8

so

\(\large 10\sqrt{36}\)

turns into

\(\large 10*6\)

I'm sure you see how to finish up

Answer 9

so what's the final answer?

Answer 10

60

Answer 11

yep, so \(\large 5\sqrt{6}*2\sqrt{6} = 60\)

Answer 12

its 2x

Answer 13

oh so not 2 times the square root of 6?

Answer 14

but 2x times the square root of 6?

Answer 15

5√6 x 2x√6

Answer 16

ok let me redo

use * to mean multiplication since x is a variable

Answer 17

\(\Large 5\sqrt{6}*2x\sqrt{6}\)

\(\Large 5*2x\sqrt{6*6}\)

\(\Large 10x\sqrt{36}\)

\(\Large 10x*6\)

\(\Large 10*6x\)

\(\Large 60x\)


So, \(\Large 5\sqrt{6}*2x\sqrt{6} = 60x\)

Answer 18

Thanks

Answer 19

yw

Answer 20

i have another one

Answer 21

ok one more

Answer 22

(4√3)(-5√21)

Answer 23

no x variables this time?

Answer 24

no

Answer 25

ok one sec

Answer 26

\(\Large (4\sqrt{3})*(-5\sqrt{21})\)


\(\Large (4)*(-5)*\sqrt{3*21}\)


\(\Large -20*\sqrt{63}\)


\(\Large -20*\sqrt{9*7}\)


\(\Large -20*\sqrt{9}*\sqrt{7}\)


\(\Large -20*3*\sqrt{7}\)


\(\Large -60*\sqrt{7}\)



In the end, \(\Large (4\sqrt{3})*(-5\sqrt{21}) = -60*\sqrt{7}\)

Answer 27

thanks

Answer 28

np