Question

4S² - 4S + 1 sum and products of the zeros​

Answer 1

Answer:

Given quadratic polynomial is

4s2 − 4s + 1

To find the zeros of the quadratic polynomial we consider

4s2 − 4s + 1 = 0

4s2 − 2s – 2s + 1 = 0

2s(2s – 1) -1(2s – 1) = 0

(2s – 1)(2s – 1) = 0

2s – 1 = 0 and 2s – 1 = 0

2s = 1 and 2s = 1

s = 1/2 and s = 1/2

∴ The zeroes of the polynomial = 1/2 and 1/2

Sum of the zeroes = -(coefficient of s)/(coefficient of s2)

Sum of the zeroes = -(-4)/4 = 1

Let’s find the sum of the roots = 1/2 + 1/2 = 1

Product of the zeros = Constant term / Coefficient of s2

Product of the zeros =1 / 4

Let’s find the products of the roots = 1/2 × 1/2 = 1/4

Step-by-step explanation:

Answer 2

Step-by-step explanation:

Given,

4s²-4s+1=0

=> 4s²-2s-2s+1 = 0

=> 2s(2s-1)-1(2s-1) = 0

=> (2s-1)(2s-1) = 0

2s-1 = 0 or 2s-1 = 0

2s = 1 or 2s = 1

s = ½ or s = ½

Let,

$\alpha = 1\2$

$\beta = 1\2$

Compare with above equation ax²+bx+c = 0

Let,

a=4 , b=-4 , c=1

Now,

Sum of the zeros=alpha+beeta = ½+½ = 1+1/2 = 2/2 = 1 =c =constant

Product of the zeros=alpha×beeta =

½×½ = ¼ = c/a = constant/coefficient of s².

I hope it helps you.