square root of 16(End square root) i^3-4i^2 all over 1- squa  TOPIC_SOLVED

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.

square root of 16(End square root) i^3-4i^2 all over 1- squa  TOPIC_SOLVED

Postby AmySaunders on Fri Sep 11, 2009 5:36 pm

K, I have a problem that I NEED help with. The square root of 16(End square root) i^3-4i^2 all over 1- square root of negative 4. I simplified it all to 4+4/1-2i. I thought that the answer was 8/1-2i. Does that sound right to you? My answer book says that the answer is 8/5+16/5i. Is that confusing to you? It is to me. Help would be appreciated. Thank you!
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Postby stapel_eliz on Fri Sep 11, 2009 6:05 pm

AmySaunders wrote:The square root of 16(End square root) i^3-4i^2 all over 1- square root of negative 4.

Does the above mean the following?

. . . . .

Is "" the imaginary, or a variable? What where the instructions? What were your steps?

Thank you! :D
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Re: square root of 16(End square root) i^3-4i^2 all over 1- squa

Postby AmySaunders on Fri Sep 11, 2009 6:07 pm

[*]square root[*]
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Re: square root of 16(End square root) i^3-4i^2 all over 1- squa

Postby AmySaunders on Fri Sep 11, 2009 6:11 pm

start square root -16 end square root i cubed minus 4 i squared all over 1 minus start square root -4 end square root.

i is an imaginary number.
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Re: square root of 16(End square root) i^3-4i^2 all over 1- squa

Postby AmySaunders on Fri Sep 11, 2009 6:16 pm

Here are my steps.

the square root of negative 16 is 4i, right?
That means that the i^3 becomes i^4,which is 2 negatives, in other words, a positive. that leaves 4. -4i^2 is also positive 4. Numerator is then 4+4,right?
in the denominator, the square root of -4 is 2i, right? If so, the denominator is 1-2i. That means the problem is 4+4/1-2i. Simplified, 8/1-2i. But the answer book says it is 8/5 + 16/5 i.

My instructions are simplify.
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Postby stapel_eliz on Fri Sep 11, 2009 6:56 pm

AmySaunders wrote:start square root -16 end square root i cubed minus 4 i squared all over 1 minus start square root -4 end square root.

i is an imaginary number.

So the expression is as follows...?

. . . . .

(Isn't this stuff just a bear to format?)

I think your steps were as follows:

. . . . .

. . . . .

. . . . .

. . . . .

. . . . .

Of course, this is not proper form, so then you rationalized the denominator by multiplying top and bottom by the conjugate of . What did you do for this step, and what did you get?
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Re: square root of 16(End square root) i^3-4i^2 all over 1- squa

Postby AmySaunders on Sat Sep 12, 2009 6:33 pm

Thank you, especially for the link! Somehow I missed learning conjugates? So I multiplied the denominator (and numerator) by the conjugate, 1+2i, and foiled, and simplified, and came up with 8/5+16i/5, which is the answer given in the answer booklet. Thank you for your help!
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