Question

4u^2 + 8u. please koei help kar do.​

Answer 1

Step-by-step explanation:

If we have to find zeros then put 4u²+8u=0

4u(u+2)=0

(4u)(u+2)=0

if 4u=0

u=0/4

u=0

if u+2=0

u=0-2

u=0

so zeros are -2 and 0

(general form of quadratic equation ax²+bx+c

where a is cofficient of x²

b is cofficient of x

and c is constant)

here a=4 and b=8 ,c=0(because nothing is given)

relationship between zeros and cofficient of quadratic equations

sum of zeros of quadratic equation=-b/a

0+(-2)= -8/4

0-2=-2

-2=-2

left hand side= right hand side

hence relationship is verified

product of zeros of quadratic equation=c/a

-2×0=0/4

0=0

Left hand side=right hand side

hence relationship is verified

it is detailed answer

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