Question

$4 {x}^{2} - 4x + 1 \: find \: th e \: zeroes$
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Answer 1

Answer:

Given,

Quadratic eq. = 4x² - 4x + 1

Here,

Coefficient of x² ( a ) = 4

Coefficient of x ( b ) = -4

Constant term ( c ) = 1

Now,

=> 4x² - 4x + 1 = 0

=> ( 2x )² - 2 × 2x × 1 + ( 1 )² = 0

Using identity :

=> ( a² - 2ab + b² ) = ( a - b )²

=> ( 2x - 1 )² = 0

=> ( 2x - 1 ) ( 2x - 1 ) = 0 ( continued further )

=> ( 2x - 1 ) = 0 ÷ ( 2x - 1 )

=> ( 2x - 1 ) = 0

=> x = 1/2

★

=> ( 2x - 1 ) = 0 ÷ ( 2x - 1 )

=> ( 2x - 1 ) = 0

=> 2x = 1

=> x = 1/2

Hence, zeroes are ( 1/2 ) and ( 1/2 ).

Now,

=> Sum of zeroes = -b/a

=> ( 1/2 ) + ( 1/2 ) = -( -4 ) ÷ 4

=> ( 1 + 1 )/2 = 4 ÷ 4

=> 2÷2 = 1

=> 1 = 1

And,

=> Product of zeroes = c/a

=> ( 1/2 ) × ( 1/2 ) = 1/4

=> 1/4 = 1/4

★ Verified ★

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Hope it helps !! ^_^

Answer 2

$\huge{\red{\boxed{\overline{\underline{\mid\mathfrak{answer}{\mathrm{\sf}\colon\mid}}}}}}$

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

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

$= > 2x(2x - 1) - 1(2x - 1) = 0$

$= > (2x - 1)(2x - 1) = 0$

$x = \frac{1}{2}$

$x = \frac{1}{2}$

HOPE IT'S HELPFUL....:-)