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FromTheMoon [43]
3 years ago
8

P(x) = x3 - 2x2 + 2x What are the real and complex zeros of the equation

Mathematics
1 answer:
vesna_86 [32]3 years ago
5 0

Given:

The polynomial is

p(x)=x^3-2x^2+2x

To find:

The real and complex zeros of the equation.

Solution:

We have,

p(x)=x^3-2x^2+2x

For zeros, p(x)=0.

x^3-2x^2+2x=0

x(x^2-2x+2)=0

x(x^2-2x+2)=0

x=0\text{ and }x^2-2x+2=0

The real value of x is 0. The equation x^2-2x+2=0 will give complex roots. Here, a=1, b=-2 and c=2.

Using quadratic formula, we get

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

x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(1)(2)}}{2(1)}

x=\dfrac{2\pm \sqrt{4-8}}{2}

x=\dfrac{2\pm \sqrt{-4}}{2}

On further simplification, we get

x=\dfrac{2\pm \sqrt{-1}\sqrt{4}}{2}

x=\dfrac{2\pm 2i}{2}

x=\dfrac{2(1\pm i)}{2}

x=1\pm i

Therefore, the real zero is 0 and the complex zeros are 1+i and 1-i.

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Un granjero tiene 100 vacas que pesan 125 kg cada una. El costo diario de mantenimiento de una vaca asciende a $ 5.00. Las vacas
solniwko [45]

Answer:

25 days

Step-by-step explanation:

Assuming that the farmer waits for n days to get get the maximum profit.

Given that the total number of cows the farmer has, = 100

Currect weight of one cow = 125 kg.

Weight gain rate for one cow = 3 kg/day

So, weight gained by one cow in n days = 3n kg

Therefore, the weight of one cow after n days = 125 + 3n \;\;kg \cdots(i)

Cost for keeping one cow = $5/day.

Cost for keeping one cow for n days = $ 5n

So, Cost for keeping 100 cows for n days = \$ 5n \times 100 =\$500n \cdots(ii)

Current market price = 25 pesos / kg = 125 cents / kg [ as 1 peso = 5 cents]

The falling rate of market price = 1 cent /day.

Fall in price in n days = 1\times n = n cents.

So, market price after n days = 125-n cent/kg\cdots(iii)

By using equations (i) and (iii),

The selling price of one cow after n days = (125+3n) \times (125-n) cents.

So, the selling price of 100 cows after n days = (125+3n) \times (125-n)\times 100 cents.

As 1 $ = 100 cents. so

The selling price of 100 cows after n days =\$ 5(125+3n) (125-n)\cdots(iv).

Now, from equations (ii) and (iv)

Net profit, P = 5(125+3n) (125-n) - 500n\cdots(v)

To get the maximum profit, differentiate the profit function in equation (v) with respect to n and equate it to zero to get the value of n, i.e

\frac {dP}{dn}=0 \\\\\Rightarrow \frac {d}{dn}(5(125+3n) (125-n) - 500n)=0 \\\\\Rightarrow  5[ (125+3n)(-1)+(125-n)(3)]-500=0 \\\\\Rightarrow 5[ -125-3n+375-3n]-500 =0 \\\\\Rightarrow 5[ -125-3n+375-3n]=500\\\\\Rightarrow 250-6n=500/5=100 \\\\\Rightarrow 6n = 250-100=150 \\\\\Rightarrow n= 150/6 = 25.

Hence, the farmer has to wait for 25 days to get the maximum profit.

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3 years ago
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lara31 [8.8K]

Answer:

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Answer:

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Solve the equation k+10=27 k=
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Steps to solve:

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