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Answer 1

Explanation:

4s

2

−4s+1

Factorize the equation, we get(2s−1)(2s−1)

So, the value of 4s

2

−4s+1 is zero when 2s−1=0,2s−1=0, i.e., when s=

2

1

or s=

2

1

.

Therefore, the zeros of 4s

2

−4s+1 are

2

1

and

2

1

.

Now,

⇒Sum of zeroes =

2

1

+

2

1

=1=−

4

−4

=−

Coefficient of s

2

Coefficient of s

⇒Product of zeros =

2

1

×

2

1

=

4

1

=

4

1

=

Coefficient of s

2

Constant term

Answer 2

Hey buddy!

Given polynomial

$= > {4x}^{2} - x - 5$

On factorising :-

$= > {4x}^{2} + 4x - 5x - 5 \\ = > 4x(x + 1) - 5(x + 1) \\ = > (x + 1)(4x - 5)$

So the zeros are -1 and 5/4

Verification of their coefficients and zeros

Sum of the zeros :-

=> coefficient of x/coffeicient of x² = (-1) + (5/4)

=> -1/4 = -1/4

And also,

Product of zeros :-

=> constant term/coefficient of x² = (-1)*(5/4)

=> -5/4 = -5/4

Hence verified!!