Biology, asked by Apeksha668, 8 months ago

Construct an aquatic food chain with 5 trophic levels.​

Answers

Answered by kaumrsatnam
8

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Attachments:
Answered by shilpapanchal126
0

Answer:

Use Vieta's method on 2x^2 + 2(m+n)x + m^2 + n^22x

2

+2(m+n)x+m

2

+n

2

.

\displaystyle\left \{ {{p + q=-(m + n)} \atop {pq =\dfrac{m^2 + n^2}{2}}} \right.

pq=

2

m

2

+n

2

p+q=−(m+n)

Two zeros of the new polynomial:

(p+q)^2=p^2+2pq+q^2(p+q)

2

=p

2

+2pq+q

2

and (p-q)^2=p^2-2pq+q^2(p−q)

2

=p

2

−2pq+q

2

Construct the new polynomial with Vieta's method.

Sum and product of the new polynomial:

Sum 2(p^2+q^2)2(p

2

+q

2

)

Product (p+q)^2(p-q)^2(p+q)

2

(p−q)

2

Finding the sum:

(p+q)^2-2pq=p^2+q^2(p+q)

2

−2pq=p

2

+q

2

=(m+n)^2-(m^2+n^2)=(m+n)

2

−(m

2

+n

2

)

=2mn=2mn

→ 4mn4mn is the sum.

Finding the product:

(p+q)^2-4pq=(p-q)^2(p+q)

2

−4pq=(p−q)

2

=(m+n)^2-2(m^2+n^2)=(m+n)

2

−2(m

2

+n

2

)

=-(m^2-2mn+n^2)=-(m-n)^2=−(m

2

−2mn+n

2

)=−(m−n)

2

→ -(m+n)^2(m-n)^2−(m+n)

2

(m−n)

2

is the product.

The new quadratic equation is x^2-4mnx-(m+n)^2(m-n)^2x

2

−4mnx−(m+n)

2

(m−n)

2

.

More information:

Vieta's Method

Consider a quadratic polynomial x^2+\dfrac{b}{a} x+\dfrac{c}{a}x

2

+

a

b

x+

a

c

.

If α and β are the zeroes of the polynomial then

(x-\alpha )(x-\beta )=x^2-(\alpha +\beta )x+\alpha \beta\;\textbf{[Factor Theorem]}(x−α)(x−β)=x

2

−(α+β)x+αβ[Factor Theorem] .

\alpha +\betaα+β is the sum of the two zeroes.

\alpha \betaαβ is the product of the two zeroes.

So \alpha +\beta =-\dfrac{b}{a}α+β=−

a

b

and \alpha \beta =\dfrac{c}{a}αβ=

a

c

.

this is answer for your upper question

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