Find the zeroes of the quadratic polynomial 4x2 ˗ 4x + 1 and verify t

1.	Find the zeroes of the quadratic polynomial 4x2 ˗ 4x + 1 and verify the relation between the zeroes and the coefficients.
Answer» 4x² ˗ 4x + 1 = 0 ⇒ (2x)² ˗ 2(2x)(1) + (1)² = 0 ⇒ (2x ˗ 1)² = 0 [∵ a² – 2ab + b² = (a–b)²] ⇒ (2x ˗ 1)² = 0 ⇒ x = \(\frac{1}2\) or x = \(\frac{1}2\) Sum of zeroes = \(\frac{1}2\) + \(\frac{1}2\) = 1 = \(\frac{1}1\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\) Product of zeroes = \(\frac{1}2\) x \(\frac{1}2\) = \(\frac{1}4\) = \(\frac{constant\,term}{(coefficient\,of\,x^2)}\)

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

Answer»

4x² ˗ 4x + 1 = 0

⇒ (2x)² ˗ 2(2x)(1) + (1)² = 0

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

⇒ (2x ˗ 1)² = 0

⇒ x = \(\frac{1}2\) or x = \(\frac{1}2\)

Sum of zeroes = \(\frac{1}2\) + \(\frac{1}2\) = 1 = \(\frac{1}1\) = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\)

Product of zeroes = \(\frac{1}2\) x \(\frac{1}2\) = \(\frac{1}4\) = \(\frac{constant\,term}{(coefficient\,of\,x^2)}\)