Question

Find the zeroes of the quadratic polynomial 4x² – 4x + 1 and verify the relation between the zeroes and the coefficients.

Find the zeroes of the polynomial 4x² – 4x + 1 and verify there rationship between the zeroes and the coefficients.

Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients:

4x² – 4x + 1

Answer 1

4x² – 4x + 1 = 0 

⇒ (2x²) – 2(2x)(1) + (1)² = 0 

⇒ (2x – 1)² = 0   ...[∵ a² – 2ab + b² = (a – b)²]

⇒ (2x – 1)² = 0 

⇒ `x = 1/2` or `x = 1/2` 

Sum of zeroes = `1/2 + 1/2`

= 1

= `1/1`

= `(-("Coefficient of"  x))/(("Coefficient of"  x^2))` 

Product of zeroes = `1/2 xx 1/2`

= `1/4`

= `("Constant term")/(("Coefficient of "x^2))`

Answer 2

Given, Polynomial is 4x² – 4x + 1   ...(i)

⇒ 4x² – 2x – 2x + 1

⇒ 2x(2x – 1) – 1(2x – 1)

⇒ (2x – 1) (2x – 1)

⇒ `x = 1/2`

Hence, zeroes of a given polynomial is `x = 1/2`

On comparing equation (i) with ax² + bx + c = 0,

We get a = 4, b = – 4 and c = 1

Now, the sum of zeroes = `(-b)/a = (-(-4))/4 = 1`

Product of zeroes = `c/a = 1/4` which matches with:

Sum of the zero = `1/2 + 1/2`

= `2/2`

= 1

Product of the zero = `1/2 xx 1/2`

= `1/4`