Question

solve the attachment​

Answer 1

Explanation:

Given :-

$\sf 4x^{2} -4x + 1 = 0$

To find :-

Relationship between the zeroes  and the  Coefficients.

Solution :-

We are knowing that

α + β = -b/a

$\sf \alpha +\beta = \dfrac{-(-4)}{1}$

$\sf \alpha +\beta = \dfrac{4}{1}$

$\alpha +\beta =4$

Product of zeroes

$\sf\alpha \beta =\dfrac{c}{a}$

$\sf \alpha \beta =\dfrac{1}{4}$

On factorizing

4x² - 4x + 1 = 0

$\sf 4x^{2} -(2x + 2x) - 1 = 0$

$\sf 4x^{2} - 2x -2x +1 = 0$

$\sf 2x(2x - 1) - 1(2x - 1) = 0.$

Taking 2x - 1 as common

$\sf (2x - 1)^2 = 0.$

$\sf x = 1/2$

Sum

x = 1/2 + 1/2

x = 1 + 1/2

x = 2/2

x = 1/1

x = 1

Product of zeroes

x = 1/2 x 1/2.

x = 1 x 1/2 x 2

x = 1/4

Answer 2

$x=1/4$

Explanation:

Let f(x) = 4x^2 ˗ 4x + 1

= (2x^2) – 2(2x)(1) + (1)^2

= (2x – 1)^2

To find the zeroes,

set f(x) = 0

(2x – 1)^2 = 0

x = 1/2 or x = 1/2

Again, Sum of zeroes

= 1/2+1/2=1=1/1

= -b/a

= (-Coefficient of x) / (Cofficient of x^2)

Product of zeroes

= 1/2 × 1/2=1/4

= c/a

= Constant term / Coefficient of x^2