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nordsb [41]
2 years ago
11

Help quickly please!

Mathematics
1 answer:
Zarrin [17]2 years ago
6 0
The answer to your question is C
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Quadratic equations and complex numbers PLEASE HELPPPP ASAP
kondor19780726 [428]

we are given

27x^3-1=0

we can also write it as

(3x)^3-(1)^3=0

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a^3-b^3=(a-b)(a^2+ab+b^2)

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we get  

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now, we can set it to 0

27x^3-1=(3x-1)(9x^2+3x+1)=0

and then we can solve for x

(3x-1)(9x^2+3x+1)=0

3x-1=0

x=\frac{1}{3}

9x^2+3x+1=0

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x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}

now, we can plug values

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x=\frac{-3\pm \sqrt{3^2-4\cdot \:9\cdot \:1}}{2\cdot \:9}

x=-\frac{1}{6}+i\frac{\sqrt{3}}{6},\:x=-\frac{1}{6}-i\frac{\sqrt{3}}{6}

So, we will get solution as

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3 years ago
(I'LL BE GIVING BRAINLIEST TO WHO EVER HELPS ME!!
motikmotik

Answer:

Step-by-step explanation:

D) six and two sixths minus three and two eighths

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2 years ago
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