What is the Square root of -72?
@ganeshie8 @girlwithsquirrels14 @Ashleyisakitty
72 = 36x2
\(\large \sqrt{-72} \) \(\large \sqrt{-36 \times 2} \)
-6 sqrt 2 6 sqrt -2 6sqrt 2i 6i sqrt 2
\(\large \sqrt{-72} \) \(\large \sqrt{-36 \times 2} \) \(\large \sqrt{-1 \times 36 \times 2} \)
Those are the options
-72 = 8.4852814
u want ansurr or learn how to do this ? :)
Learn how please
u sure ? dont lie
Yeah I want to understand!
pull the 36 out, it becomes 6 ?
good :) \(\large \sqrt{-72} \) \(\large \sqrt{-36 \times 2} \) \(\large \sqrt{-1 \times 36 \times 2} \) \(\large 6\sqrt{-1 \times 2} \)
see if above steps make sense
Eh not really ! I dont know this
\(\large \sqrt{-72} \) \(\large \sqrt{-36 \times 2} \)
till that step u okays ? :)
yess!
since, sqrt(36) = 6 it becomes :- \(\large 6 \sqrt{-1 \times 2} \)
remember this for the rest of ur course time : sqrt(-1) = i so it becomes :- \(\large 6 i \sqrt{2} \)
Oh okay . And what are the steps for solving 2 divided by 2 + 5i?
@ganeshie8
multiply top and bottom wid conjugate
\(\large \frac{2}{2+5i} \) multiply top and bottom wid conjugate 2-5i \(\large \frac{2}{2+5i} \times \frac{2-5i}{2-5i}\)
wait so 4 and 10i
\(\large \frac{2}{2+5i} \) multiply top and bottom wid conjugate 2-5i \(\large \frac{2}{2+5i} \times \frac{2-5i}{2-5i}\) \(\large \frac{2(2-5i)}{(2+5i)(2-5i)}\) since, \((a+ib)(a-ib) = a^2+b2, \) \(\large \frac{2(2-5i)}{2^2+5^2}\) \(\large \frac{4-10i}{29}\)
Wow thats tricky for me ..I see what you mean tho . thank you !
glad u cud make some sense of that :) yw !
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