Can anyone help me with these..?!

1) 3i/2+5i

and...

2) square root of -2(square root of -9 - square root of 7)

i would appreicate it if you can reply asap.

P.S: only reply if u KNOW the answer...thanks.

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- Jan 22nd 2006, 11:22 AM #1

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## What the?

Can anyone help me with these..?!

1) 3i/2+5i

and...

2) square root of -2(square root of -9 - square root of 7)

i would appreicate it if you can reply asap.

P.S: only reply if u KNOW the answer...thanks.

- Jan 22nd 2006, 11:49 AM #2

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Originally Posted by**MOODYtarqi**

but I will presume you are asking for:

$\displaystyle \frac{3i}{2+5i}$

to be reduced to standard form $\displaystyle a+b.i$.

Multiply the top and bottom by the complex conjugate of the bottom:

$\displaystyle \frac{3i}{2+5i} = \frac{3i (2-5i)}{(2+5i)(2-5i)}$,

so:

$\displaystyle \frac{3i}{2+5i} = \frac{15+6i}{29}=\frac{15}{29}+\frac{6i}{29}$.

RonL

- Jan 22nd 2006, 11:51 AM #3

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- Jan 22nd 2006, 12:02 PM #4

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Originally Posted by**MOODYtarqi**

$\displaystyle \sqrt{-2(\sqrt{-9}-\sqrt{7}})$,

or:

$\displaystyle \sqrt{-2}(\sqrt{-9}-\sqrt{7})$.

I will assume that you mean the latter of these, then:

$\displaystyle \sqrt{-2}(\sqrt{-9}-\sqrt{7})=\sqrt{2}i(3i-\sqrt{7})$

so:

$\displaystyle \sqrt{-2}(\sqrt{-9}-\sqrt{7})=-3\sqrt{2}-\sqrt{14}i$

RonL

- Jan 22nd 2006, 12:04 PM #5

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- Jan 22nd 2006, 12:10 PM #6

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- Jan 22nd 2006, 12:12 PM #7

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