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

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
AmySaunders
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### square root of 16(End square root) i^3-4i^2 all over 1- squa

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!

stapel_eliz
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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?

. . . . .$\frac{\sqrt{16}i^3\, -\, 4i^2}{1\, -\, \sqrt{-4}}$

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

Thank you!

AmySaunders
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### Re: square root of 16(End square root) i^3-4i^2 all over 1- squa

[*]square root[*]

AmySaunders
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### Re: square root of 16(End square root) i^3-4i^2 all over 1- squa

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.

AmySaunders
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### Re: square root of 16(End square root) i^3-4i^2 all over 1- squa

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.

stapel_eliz
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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...?

. . . . .$\frac{\sqrt{-16}\,i^3\, -\, 4i^2}{1\, -\, \sqrt{-4}}$

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

I think your steps were as follows:

. . . . .$\frac{(4i)(i^3)\, -\, 4(-1)}{1\, -\, 2i}$

. . . . .$\frac{4i^4\, +\, 4}{1\, -\, 2i}$

. . . . .$\frac{4(-1)^2\, +\, 4}{1\, -\, 2i}$

. . . . .$\frac{4\, +\, 4}{1\, -\, 2i}$

. . . . .$\frac{8}{1\, -\, 2i}$

Of course, this is not proper form, so then you rationalized the denominator by multiplying top and bottom by the conjugate of $1\, -\, 2i$. What did you do for this step, and what did you get?

AmySaunders
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### Re: square root of 16(End square root) i^3-4i^2 all over 1- squa

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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