Question

What am I doing wrong?
Preform the indicated operation. Assume all variables represent nonnegative real numbers

Answer 1

\[\sqrt{6x ^{2}}+x \sqrt{52}\]

Answer 2

use the product rule to simplify the first radical \[\sqrt{6x ^{2}}=\sqrt{x ^{2}}\times \sqrt{6}=x \sqrt{6}\]

Answer 3

For the second radical factor 54
\[x \sqrt{54}=x \sqrt{9\times6}\]

Answer 4

use the product rule
\[x \sqrt{9\times6}=x \sqrt{9}\times \sqrt{6}\]
\[x \sqrt{9\times6}=3x \sqrt{9}\]

Answer 5

you mean \(3x \sqrt 6\)
right?

Answer 6

combine both radicals \[(9x+3)\sqrt{9}=3x \sqrt{9}\]

Answer 7

also
6*9 is not 52

Answer 8

oh just a typo

Answer 9

It's 54 I messed up

Answer 10

\(x\sqrt 6 + 3x\sqrt 6 = (x+3x)\sqrt 6 = 4x \sqrt 6 \)

Answer 11

\(\color{blue}{\text{Originally Posted by}}\) @ScarlettFarra2000
use the product rule
\[x \sqrt{9\times6}=x \sqrt{9}\times \sqrt{6}\]
\(x \sqrt{9\times6}=3x \sqrt{9}\) <<<<<<
\(\color{blue}{\text{End of Quote}}\)

\(x \sqrt{9\times 6} = x \sqrt 9 \sqrt 6 = 3x\sqrt 6 \)

Answer 12

Oh so its not \[3x \sqrt{9}\] like i thought its \[3x \sqrt{6}\]

Answer 13

\(\Large x \sqrt{9\times 6} = x \sqrt 9 \sqrt 6 = 3x\sqrt 6 \)

^^

Answer 14

doubt cleared? any more doubts?

Answer 15

Well when I put the answer in on my assignment to still tells me I'm wrong

Answer 16

\(4x \sqrt 6\) is correct...

Answer 17

Yes thank you so much

Answer 18

welcome ^_^