Factorise the following using appropriate identities-9 y 2-6 y -1A- 4 y-13 y-1B- 7 y-13 y-1C- 3y-1 3y-1D- 2 y-13 y-1

The correct option is C (3y 1) (3y 1) 9y^2 – 6y + 1= (3y)^2 – 2(3y)(1) + 1^2 we have (a b)^2 = a^2 – 2ab + b^2 ⇒(3y 1)^2 =(3y)^2 – 2(3y)(1) + 1^2 OR (3y) ...

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Standard X

Mathematics

Quadratic Equation as Product of 2 First Degree Equations

Factorise the...

Question

Factorise the following using appropriate identities:
$9 y^{2} - 6 y + 1$

(3y-1) (3y-1)

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(2y-1) (3y-1)

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(4y-1) (3y-1)

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(7y-1) (3y-1)

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Solution

The correct option is A
(3y-1) (3y-1)

$9 y^{2} - 6 y + 1 = {(3 y)}^{2} - 2 (3 y) (1) + 1^{2}$

we have ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$

$\Rightarrow {(3 y - 1)}^{2} = {(3 y)}^{2} - 2 (3 y) (1) + 1^{2}$

OR

${(3 y)}^{2} - 2 (3 y) (1) + 1^{2} = {(3 y - 1)}^{2} = (3 y - 1) (3 y - 1)$

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Factorise the following using appropriate identities: 9y2–6y+1

The correct option is A (3y-1) (3y-1) 9y2–6y+1=(3y)2–2(3y)(1)+12 we have (a−b)2=a2–2ab+b2 ⇒(3y−1)2=(3y)2–2(3y)(1)+12 OR (3y)2–2(3y)(1)+12=(3y−1)2=(3y−1)(3y−1)

Factorise the following using appropriate identities:
$9 y^{2} - 6 y + 1$

The correct option is A
(3y-1) (3y-1)

$9 y^{2} - 6 y + 1 = {(3 y)}^{2} - 2 (3 y) (1) + 1^{2}$

we have ${(a - b)}^{2} = a^{2} - 2 a b + b^{2}$

$\Rightarrow {(3 y - 1)}^{2} = {(3 y)}^{2} - 2 (3 y) (1) + 1^{2}$

OR

${(3 y)}^{2} - 2 (3 y) (1) + 1^{2} = {(3 y - 1)}^{2} = (3 y - 1) (3 y - 1)$