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Multiply and simplify Radical Expression 1/2√5(1 1/2√20)

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Hi I got the answer of 7/2√5 but when I do it I get something different.
okay so lets write that a little prettier to deal with:\[{{1\over2}\sqrt{5}}\times{1~{1\over2}\sqrt{20}}~~\implies~~{{1\over2}\sqrt{5}}\times{{3\over2}\sqrt{20}}~~\implies~~{\sqrt{5}\over2}\times{3\sqrt{20}\over2}\]now that that looks a little better, let's solve:\[{{\sqrt{5}\times3\sqrt{20}}\over2}={3\sqrt{100}\over2}\]because 5 x 20 = 100\[3(10)\over2\]i think you've got it from here :)

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Other answers:

oh i meant to write \[3(10)\over4\]sorry
I'm a bit lost I have to simplify this so would'nt 1 1/2√20 be 2 3/2√5
\[1 {1\over2} = {3\over2}\]\[\sqrt{20}= 2\sqrt{5}\]\[{3\over2}\times{2\over1}\sqrt{5}~~=~~{3\over\cancel{2}}\times{\cancel{2}\over1}\sqrt{5}~~=~~\large3\sqrt{5}\]
but the answer comes out to \[\frac{ 7 }{ 2 }\sqrt{5}\]
and where are you getting this from?
Our teacher placed the answer as \[\frac{ 7 }{ 2 }\sqrt{5}\]
I didn't get this answer either this is why it's confusing
1/2 times 3/2 = 3/4 sqrt 5 times sqrt 20 = sqrt100 = 10 3/4 times 10 = 7.5 i dont know what your teacher was doing, maybe you copied the question wrong
No it's printed on my paper
then im really not sure
I appreciate your help Thank you yummydum
no problem :)
I got as my final answer \[\frac{ 3 }{ 4 }\sqrt{5} \] this is why it was confusing
I had a person help me don't I don't believe it's correct here is why unless I'm wrong
Alright so i read the whole problem. Where are you stuck? We should get 15/2.
\[1\frac{ 1 }{ 2 }\sqrt{20}\] would be \[\frac{ 3 }{ 2 }\] and \[\sqrt{20} would be 2\] add both \[2\frac{ 3 }{ 2 }\sqrt{5}\]
teacher has an answer of \[\frac{ 7 }{ 2 }\sqrt{5}\] could the teacher be wrong
Yes. Teacher added the sqrt5 by misatke. Or maybe the question you wrote is incorrect.
question is multiply and simplify \[\frac{ 1 }{ 2 }\sqrt{5}\left( 1\frac{ 1 }{ 2 } \sqrt{20}\right)\]
Why is this so complicated? I'm lost
Which step is it that you don't understand?
how would I multiply \[\frac{ \sqrt{5} }{ 2 }\times \frac{ 3*2 }{ 2 }\sqrt{5}\] I thought you multiply numbers not square roots with numbers
my answer was \[\frac{ 3 }{ 4 }\sqrt{5}\] maybe this is wrong
Wouldn't we keep the root? and just multiply getting \[\frac{ 15 }{ 2 }\sqrt{5}\]
No ma'am. Whenever it's given: sqrt3 * sqrt3 we will get "3" as answer. The roots cancel out leaving you with ONE three.
This is how I started \[\frac{ 1 }{ 2 }\sqrt{5}(1\frac{1 }{ 2 }\sqrt{20})\] \[\frac{ 1 }{ 2 }\sqrt{5}(\frac{3 }{ 2 }\sqrt{5})\] \[\frac{ 1 }{ 2 }\times \frac{ 3 }{ 2 }=\frac{ 3 }{ 4 }\sqrt{5}\] I know this isn't corrected but I get this
How will you simplify sqrt 20?
that's right it would be 4*5 which 4 goes down to 2 so it would be \[2\sqrt{5}\]
I forgot to add that \[\frac{ 1 }{ 2 }\sqrt{5}(2\frac{3 }{ 2 }\sqrt{5})\] it would be this correct and mix fractions to improper fractions would be \[2\frac{ 3 }{ 2 } would be \frac{ 7 }{ 2 }\] \[\frac{ 1 }{ 2 }\sqrt{5}(\frac{ 7 }{ 2 }\sqrt{5})\]
You made a mistake again. :P It's not 2 3/2
so it's \[\frac{ 2 }{ 1 }X\frac{ 3 }{ 2 }\]
No. I think you should redo the problem. Show me yor steps please.
\[\frac{ 1 }{ 2 }\sqrt{5}(1\frac{ 1 }{ 2 }\sqrt{20})\] \[\frac{ 1 }{ 2 }\sqrt{5}(\frac{ 3 }{ 2 } 2\sqrt{5})\] \[\frac{ 1 }{ 2 }\sqrt{5}(\frac{ 3 }{ 2 } 2\sqrt{5})\] at this point I'm lost
Awesome. No the 2's on right cancels out. Right?
why would it cancel √20 would be 2√5 this is where I'm getting confused
is 2 canceling with the other 2
sqrt(20) = sqrt(2*2*5) = 2 sqrt5 Agree?
Now the 2 from 2sqrt5 and the w in the denominator on the right canel out
I'm with you on that
Right o what are we left with now?
Right. And that is multiplied with..?
\[\frac{ 1 }{ 2 }\sqrt{5}\]
Right now multiply it out.
\[\frac{ 1 }{ 2 }\sqrt{5}\times \frac{ 3 }{ 1 }\sqrt{5}\] Is this looking correct?
\[\frac{ 3 }{ 2 }\sqrt{5}\]
do I take of the √ and make it \[\frac{ 3 }{ 2 }\times \frac{ 5 }{ 1 }=\frac{ 15 }{ 2 }\]
Like a boss. (y)
Thanks for everything wow :.)
Good job helping @newatthis !!! @saifoo.khan
Kudos to @yummydum too!
@yummydum is useless.
No she isn't :p
I'M useless.
Yummydum isn't. Or you,for that matter.
Thanks for everything it helps a lot when someone knows how to guide and explain it. :)
@help123please. believe me. I know her. @newatthis You're welcome. :)
Well,'ve been here longer than me.
Lol. But no doubt she has a brown brain like me! D:
Yours is a great brain,Saif.
And it's brown too!
Nope it's not.
I'm brown. Brain is brown too. Dummy is brown too. Hence she has a brown brain too.
We're prolly weirding @newatthis out. Or at least,I am.
'Hence,she has a brown brain too'. Hence. I like that word.
no I'm just doing math working my brain, or whatever is left. :)
Saif,I just noticed your smartscore is mine,with the numbers flipped upside down. XD
Haha. Nice catch.
*Flips PC upside down* OMGEE I HAVE A 99 SS!
98% of the time,when typing lol,you're not actually laughing. Did u know? :P
Make that 99%. I love laughing. It helps in exercising my face muscles. :P
so new question? \[2x \sqrt{7}(3x \sqrt{7})\] \[2x(3x)\sqrt{7}\sqrt{7}\] \[6x(\sqrt{49})\] Would this be correct?
sorry I just threw this in here
@newatthis please close this post and create a new thread.
2x * 3x = 6x^2 not just 6x and √49 = 7 so now you have 6x^2 * 7 ....which is?
\[42x ^{2}\]

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