Question

find a if
sqrta(sqrt6)=3sqrt2

find b if
sqrt10(sqrt5+sqrtb)=5sqrt2+2sqrt5

Answer 1

\[\large \sqrt{\sqrt{6}}=3\sqrt{2}\]
Like this? D:

Answer 2

no

Answer 3

\[\large \sqrt{a}\sqrt{6}=3\sqrt{2}\]\[\large \sqrt{10}(\sqrt{5}+\sqrt{b})=5\sqrt2+2\sqrt5\]
Like this?

Answer 4

\[Find a if \sqrt{a}(\sqrt{6})=3\sqrt{2}\]

Answer 5

yes like that

Answer 6

Alright \[\large \sqrt{a}\sqrt{6}=3\sqrt{2}=\sqrt{9}\sqrt{2}=\sqrt{2*9}=\sqrt{18}\]
\(\large \sqrt{a}\sqrt{6}=\sqrt{18}\rightarrow \sqrt{a}=\frac{\sqrt{18}}{\sqrt{6}}\rightarrow a=\sqrt{\frac{18}{6}}^2=\sqrt{3}^2=3\)

Answer 7

i dont get it

Answer 8

how did u know a was 9

Answer 9

nvm i get that

Answer 10

i dont get the last part

Answer 11

Which part?

Answer 12

The second line?

Answer 13

yes

Answer 14

is the answer not sqrt18

Answer 15

That was still the first question!

Answer 16

its ok nvm i understand, what about the next one

Answer 17

do i do the f2nd one like the first

Answer 18

b=2 correct??

Answer 19

\[\large \begin{align}
\sqrt{10}(\sqrt{5}+\sqrt{b})&=5\sqrt2+2\sqrt5 \\
\sqrt{10*5}+\sqrt{10b}&=\sqrt{25}\sqrt{2}+\sqrt{4}\sqrt{5}\\
\sqrt{50}+\sqrt{10b}&=\sqrt{50}+\sqrt{20}\\
\sqrt{50}+\sqrt{10b}-\sqrt{50}&=\sqrt{20}\\
\sqrt{10}*\sqrt{b}&=\sqrt{20}\\
\sqrt{b}&=\sqrt{\frac{20}{10}}\\
\sqrt{b}&=\sqrt{2}\\
b&=2

\end{align}\]

Answer 20

Correct! :)

Answer 21

thanks