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harina [27]
3 years ago
8

2x^2-3x+1=0, how do I solve this

Mathematics
1 answer:
cricket20 [7]3 years ago
7 0
<span>2x^2-3x+1=0
(2x -1)(x-1) = 0
2x - 1 = 0; x = 1/2
x - 1 = 0;x = 1

answer
</span>x = 1/2 and x = 1
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Answer:

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Step-by-step explanation:

The formula for the volume of a cylinder is

V = πr²x

Data:

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Calculations:

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Divide each side by π

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1024 =    x³ - 16x² + 64x

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2. Solve the cubic equation

The general formula for a third-degree polynomial is

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Your polynomial is  

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According to the <em>rational roots theorem</em>, the possible roots are

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Factors of d = ±1, ±2, ±4, ±8, ±16, ± 32, ± 64, ±128, ±256, ±512, ±1024

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That's a lot of possibilities to check by trial and error. I will just use the one that works.

Try x = 16 by synthetic division.

16|1  -16   64  -1024

  <u>|     16     0   1024 </u>

   1     0   64        0

So,

(x³ - 16x² + 64x - 1024)/(x – 16) = x² + 64

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(x - 16)(x² + 64) = 0

x - 16 = 0        x² + 64 =    0

     x = 16       x²         = -64

                       x          =  ±8i

There is only one real root: x = 16 mm

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