Question

4x² - 4x +1 find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients​

Answer 1

We have,

4x² - 4x + 1 = 0 [ Given ]

⇒4x² - 2x - 2x + 1 = 0

⇒ 2x ( 2x - 1 ) - 1 ( 2x - 1 ) = 0

⇒ ( 2x - 1 ) ( 2x - 1 ) = 0

⇒ 2x - 1 = 0 and 2x - 1 = 0

⇒ x = 1/2 and x = 1/2

Therefore, the zeroes of 4x² - 4x + 1 are 1/2 and 1/2.

★ Sum of the zeroes :

1/2 + 1/2 = - coefficient of x / coefficient of x²

⇒ 1 + 1 / 2 = - ( - 4 ) / 4

⇒ 2/2 = 4/4

⇒ 1 = 1

★ Product of the zeroes :

1/2 × 1/2 = constant term / coefficient of x²

⇒ 1/4 = 1/4

Verified.

Answer 2

✯✯ QUESTION ✯✯

4x² - 4x +1 find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients..

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✰✰ ANSWER ✰✰

$\longmapsto\tt\bold{{4x}^{2}-4x+1}$

By Splitting Middle Term : -

$\longmapsto\tt\bold{{4x}^{2}-4x+1=0}$

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

$\longmapsto\tt{2x(2x-1)-1(2x-1)}$

$\longmapsto\tt\bold{(2x-1)(2x-1)}$

• x = 1/2
• x = 1/2

➮So , 1/2 and 1/2 are the zeroes of polynomial 4x²-4x+1...

➥Here : -

• a = 4
• b = -4
• c = 1

★Sum of Zeroes : -

$\longmapsto\tt{\alpha+\beta=\dfrac{-b}{a}}$

$\longmapsto\tt{\dfrac{1}{2}+{1}{2}=\dfrac{-(-4)}{4}}$

$\longmapsto\tt{\cancel\dfrac{2}{2}=\cancel\dfrac{4}{4}}$

$\longmapsto\tt\bold{1 = 1}$

$\red\longmapsto\:\large\underline{\boxed{\bf\green{L.H.S}\orange{=}\purple{R.H.S}}}$

★Product Of Zeroes : -

$\longmapsto\tt\bold{\alpha\beta=\dfrac{c}{a}}$

$\longmapsto\tt{\dfrac{1}{2}\times{\dfrac{1}{2}}=\dfrac{1}{4}}$

$\longmapsto\tt\bold{\dfrac{1}{4}=\dfrac{1}{4}}$

$\pink\longmapsto\:\large\underline{\boxed{\bf\blue{R.H.S}\red{=}\orange{R.H.S}}}$