Question

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Answer 1

4u^2 +8u

to calculate the zeros of given equation

put f (u) =0

4u^2 +8u=0

4u(u+2)=0

the zeros of the given equation are u=0, u=-2

sum of zero is 0+(-2)=-2

product of the zeros is 0×-2=0

according to the given equation

The sum of zero is

-b/a =-8/4

=-2

the product of zero is

c/a= 0/4

hence it is verified that

sum of zeros = -coefficent of x/ coefficient of x^2

And,

product of zero = constant term/ coefficient of x^2

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