SOLUTION: This is my last Algebra 2 problem and I have no idea how to solve this. The problem is to simplify -1/(3+i) using imaginary numbers. To me it appears that this is as simplified
Question 275548: This is my last Algebra 2 problem and I have no idea how to solve this. The problem is to simplify -1/(3+i) using imaginary numbers. To me it appears that this is as simplified as it can get.
Thank you.
Mark Found 2 solutions by stanbon, richwmiller:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! simplify -1/(3+i) using imaginary numbers
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"Simplified" in this case means "no imaginary numbers in the denominator".
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So, multiply numerator and denominator by the conjugate of 3+i, which is 3-i.
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You get:
[-1(3-i)]/[(3+i)(3-i)]
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The form in the denominator is (a+b)(a-b)
That product is always a^2 - b^2
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So for your problem you get:
[(-3+i)]/[3^2-(i)^2]
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= (-3+i)/(9 + 1)
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= (-3+i)/10
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That is the simplified form.
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Cheers,
Stan H.
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