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avatar+449 

Hi good people!, for the following sum I get a different answer to the book, and no matter how many times I do this sum, I keep getting the same answer. please help me to see where I'm going wrong?

 

\(\sqrt{-5}(\sqrt{-3} * \sqrt{-4})\)

\(\sqrt{5*-1}(\sqrt{3*-1}* \sqrt{2^2 *-1})\)

\(\sqrt{5i}(\sqrt{3i}* \sqrt{2^2i})\)

\(\sqrt{5i}(\sqrt{3i}* 2 \sqrt{i})\)

\(\)\(\sqrt{15i^2}*2(-1)\)\(​​​​\)\(​​​​\)

\(-2 \sqrt{15i^2}\)

 

The book says \(-2 \sqrt{15i}\)

 

Thanx for the help!!

 Jan 22, 2019
 #1
avatar+101741 
+3

Hi Juriemagic :)

 

\(\sqrt{-5}=\sqrt{5*-1}=\sqrt5*\sqrt{-1}=\sqrt5\;i\)

 

So your mistakes strat in the third row.

 

\(\sqrt{-5}(\sqrt{-3}*\sqrt{-4})\\ \sqrt5*\sqrt3*\sqrt4*i*i*i\\ 2\sqrt{15}*-1*i\\ -2\sqrt{15}\;i\)

 

 

This is different from your answer but different from your book too.  

Are you sure you copied in the book answer correctly?

 Jan 22, 2019
 #4
avatar+449 
+1

Hello Melody!!!!,

 

I am going to be VERY honest about this....the answer I actually got, was exactly the same as what you got...I promise you that. this morning when I posted the problem, i did not have my paper that i worked on, with me, so I did the sum afresh...and got lost. I figured that it was'nt a train smash if i did not put the sum exactly as I actually had it....the answer is wrong any case...so I left it as such

 

Yes I also did the 3 "i"'s etc...

Melody, yes, I checked and re-checked that answer....the "i" looks like it is within the square root sign...maybe they did a mis-print?

juriemagic  Jan 22, 2019
 #5
avatar+101741 
+1

Yes it would have been a misprint.  :)

 

or a poor print  ;)

Melody  Jan 22, 2019
edited by Melody  Jan 22, 2019
 #6
avatar+449 
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Melody,

Thank you very much for your assistance...appreciated!!

juriemagic  Jan 22, 2019
 #7
avatar+101741 
+1

It is always a pleasure to help you Juriemagic     wink

Melody  Jan 22, 2019

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