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\[a ^{2}\sqrt{8} /\sqrt{16a^{2}}\]
\[\sqrt{8} = \sqrt{4}*\sqrt{2}\] \[\sqrt{16a^{2}} = \sqrt{16}*\sqrt{a^{2}}\] that should help
not really lol sorry

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Other answers:

what is sqrt(4) ?
what??
do you know what the the square root of 4 is?
ok how about 16?
2 and 4...
yep, now a^2
@dumbcow u lost me a long time ago
\[\sqrt{16} = \sqrt{4^{2}} = 4\] so for any number \[\sqrt{a^{2}} = a\]
lol ur not hepling my question tho...
well to simplify the problem , you need to simplify or get rid of the radicals, that is what im trying to help you do
dont u just multiply the top by the bottom???
you want someone to answer your problem? find the square root of 14 of a^2 and sqrt(4)*sqrt(2)
\[\frac{\sqrt{8} a^{2}}{\sqrt{16a^{2}}} = \frac{\sqrt{4}\sqrt{2}a^{2}}{\sqrt{16}\sqrt{a^{2}}} = \frac{2\sqrt{2}a^{2}}{4a}\]
@dumbcow ok now i seee ima sorry about that
np :) now just reduce to get final answer
lol thanks!!

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