Question

4u²+8u


Find the zeros of the following quadratic polynomial and verify the relationship between the zeros of the coefficient

Answer 1

4u^2 + 8u

Take 4u common

4u(u+2)

Let 4u =0

U =0

Similarly u+2=0

U=-2

Verification

Alpha+ bheta = (-) coff.ofu÷coff.ofu^2

=-8÷4=-2

Alpha×bheta = constant÷coff.ofu^2

0÷4 =0

Hence verified

Answer 2

Here your answer goes

Step :- 1

Given ,

$4u^2 + 8u$

Take 4 Common it will $4u ( u + 2 )$

Step :- 2

So obtained zeroes are 0 , -2

Step :- 3

We know ,

Sum of the zeroes = $\alpha + \beta = \frac{-b}{a}$

Product of the zeroes = $\alpha * \beta = \frac{c}{a}$

Step :- 4

Put the values

Sum of zeroes = 0 + 2 = -2 = $\frac{-8}{4}$

Product of Zeroes =  0 * (-2) = 0 = $\frac{0}{4}$

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