Question

Help with simplifying radical expressions??? fan/medal

Answer 1

\[\sqrt{\frac{ 108x ^{13}y ^{3} }{ 27xy ^{7} }}\]

Answer 2

So far I have:

\[\frac{ 6\sqrt{3x ^{13}y ^{3}} }{ ? }\]

Answer 3

Am I not simplifying x and y correctly?

Answer 4

\[\large \dfrac{a^m}{a^n} = a^{m-n} = \dfrac{1}{a^{n-m}}\]

Answer 5

Notice that \(108 = 27\times 4\)

Answer 6

\[\sqrt{\frac{ 108x ^{13}y ^{3} }{ 27xy ^{7} }} = \sqrt{\frac{27\times 4 x^{13}y^3}{27xy^7}}\]

Answer 7

wait now I'm confused

Answer 8

I've never learned it this way

Answer 9

isnt it easy this way ?

\[\require{cancel} \sqrt{\frac{ 108x ^{13}y ^{3} }{ 27xy ^{7} }} = \sqrt{\frac{27\times 4 x^{13}y^3}{27xy^7}} = \sqrt{\frac{\cancel{27}\times 4 x^{13}y^3}{\cancel{27}xy^7}} = \sqrt{\frac{4 x^{13}y^3}{xy^7}} \]

Answer 10

Ok thats easy. But what if it the numbers didn't multiply into each other like that?

Answer 11

then simplification is not possible :)
lets keep going with the present problem and finish it off

Answer 12

Ok

Answer 13

you have x^13 in numerator and a x in denominator
what does that simplify to ?

Answer 14

x^12

Answer 15

Excellent!
what about y^3/y^7 ?

Answer 16

y^-4 ?

Answer 17

Right! but we dont want negative exponents so can say it becomes 1/y^4 ?

Answer 18

So do we just move it to the bottom?

Answer 19

\[\begin{align}\require{cancel} &\sqrt{\frac{ 108x ^{13}y ^{3} }{ 27xy ^{7} }} = \sqrt{\frac{27\times 4 x^{13}y^3}{27xy^7}} = \sqrt{\frac{\cancel{27}\times 4 x^{13}y^3}{\cancel{27}xy^7}} = \sqrt{\frac{4 x^{13}y^3}{xy^7}}\\~\\~\\
&= \sqrt{\frac{4 x^{13-1}}{y^{7-3}} }= \sqrt{\frac{4x^{12}}{y^4}}
\end{align}\]

Answer 20

wid me so far ?

Answer 21

yep! :)

Answer 22

good write them as powers of 2 and kill the sqrt

Answer 23

would the 4 simplify to 2?

Answer 24

yes!
\[\begin{align}\require{cancel} &\sqrt{\frac{ 108x ^{13}y ^{3} }{ 27xy ^{7} }} = \sqrt{\frac{27\times 4 x^{13}y^3}{27xy^7}} = \sqrt{\frac{\cancel{27}\times 4 x^{13}y^3}{\cancel{27}xy^7}} = \sqrt{\frac{4 x^{13}y^3}{xy^7}}\\~\\~\\
&= \sqrt{\frac{4 x^{13-1}}{y^{7-3}} }= \sqrt{\frac{4x^{12}}{y^4}} =
\sqrt{\frac{2^2(x^6)^2}{(y^2)^2} } = \sqrt{\frac{(2x^6)^2}{(y^2)^2}} = \frac{2x^6}{y^2}
\end{align}\]

Answer 25

Ok thats what i got :)

Answer 26

Could you possibly help me with another one like this?

Answer 27

wil try, ask..

Answer 28

\[\frac{ \sqrt{50x ^{10}}y ^{5} }{ \sqrt{98x ^{2}}y ^{15}}\]

Answer 29

y terms are outside the radical is it ?

Answer 30

?

Answer 31

\(y^5\).. is it really outside the radical ?

Answer 32

no its not

Answer 33

it's inside

Answer 34

\[\large \frac{ \sqrt{50x ^{10}y^5} }{ \sqrt{98x^2y ^{15}}}\]

like this ?

Answer 35

Yep sorry

Answer 36

put them in a single radical

Answer 37

you are allowed to do that as everything is real

Answer 38

\[\large \frac{ \sqrt{50x ^{10}y^5} }{ \sqrt{98x^2y ^{15}}} = \sqrt{\frac{50x ^{10}y^5}{98x^2y ^{15}}}\]

Answer 39

looks 2 goes in both 50 and 98 ?
cancel it

Answer 40

\[\frac{ 5x ^{8} }{ 7y ^{10}}\]

Answer 41

since 25 and 49 can both be squared

Answer 42

careful, you're right about 5/7 but you need to take sqrt of variables also

Answer 43

do you do that before you subtract them? do you always subtract them?

Answer 44

\[\frac{ \sqrt{50x ^{10}y^5} }{ \sqrt{98x^2y ^{15}}} = \sqrt{\frac{50x ^{10}y^5}{98x^2y ^{15}}} = \sqrt{\frac{25x ^{10}y^5}{49x^2y ^{15}}} = \frac{5}{7}\sqrt{\frac{x^{10}y^5}{x^2y^{15}}}\]

fine so far ?

Answer 45

Oh and you divide the variables I forgot for some reason

Answer 46

yes try it again

Answer 47

what do you do with y^5 and y^15 that can't be divided by 2?

Answer 48

simplify first

Answer 49

\[\frac{ 5x ^{4} }{ 7y ^{5} }\]

Answer 50

Yep!
\[\begin{align} &\frac{ \sqrt{50x ^{10}y^5} }{ \sqrt{98x^2y ^{15}}} = \sqrt{\frac{50x ^{10}y^5}{98x^2y ^{15}}} = \sqrt{\frac{25x ^{10}y^5}{49x^2y ^{15}}} = \frac{5}{7}\sqrt{\frac{x^{10}y^5}{x^2y^{15}}}\\~\\~\\
&\frac{5}{7}\sqrt{\frac{x^{10-2}}{y^{15-5}}} =\frac{5}{7}\sqrt{\frac{x^{8}}{y^{10}}} = \frac{5x^4}{y^5}
\end{align}\]

Answer 51

Thank you so much! I understand it now.

Answer 52

you're welcome !