Question

find the zero of 4s power2 - 4s +1 and ve​

Answer 1

Answer:

Step-by-step explanation:

given ,

quadratic polynomial = 4s²- 4s +1= q(y)

to find the zeroes of the given quadratic polynomial,

q(y) = 0

4s² - 4s+1 = 0

⇒4s²-2s-2s+1 = 0

⇒2s(2s-1)-1(2s-1) = 0

⇒(2s-1) (2s-1) = 0

⇒2s - 1  = 0 ; 2s - 1= 0

⇒s = 1/2, 1/2

therefore, the zeroes of the polynomial = 1/2,1/2

now,

sum of the zeroes = 1/2 + 1/2 = 2/2 = -(co-efficient of s)

                                                             (co-efficient of s²)

product of the zeroes = 1/2(1/2) = 1/4 = constant term / coefficient of s²

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Answer 2

Step-by-step explanation:

The question is to find the root of 4s² - 4s + 1 = 0 I guess because the question is invalid.

So, it'll be hard for you to do by factorisation method.

Well use quadratic formula to do this.

note that a = 4; b = -4; c = 1

The formula is

$\frac{ - b \: \pm \: \sqrt{ {b}^{2} - 4ac} }{2a} \\ \\ \\ putting \: the \: values \\ \\ \\ \frac{4 \: \pm \: \sqrt{ {4}^{2} - 4 \times 4 \times 1} }{2 \times 4} \\ \\ \\ \frac{4 \: \pm \: \sqrt{16 - 16} }{8} \\ \\ \\ \frac{4}{8}$

Therefore the value of s is 4/8.