Find the value of nbsp- -3a-1-2-3a-1-2-9a-2-2a-1-4-9a-2-3a-1-4-6a-2-3a-1-4-9a-2-3a-1-2-

The correct option is B 9a^2 3a+1/4 (3a 1/2)(3a 1/2) =(3a 1/2)^2 Using the identity nbsp; (a b)^2=a^2 2ab+b^2, (3a 1/2)^2 =(3a)^2 2× 3a×1/2+(1/2)^2 =9a^ ...

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Byju's Answer

Standard VII

Mathematics

Multiplication of a Binomial by a Binomial

Find the valu...

Question

Find the value of $(3 a - \frac{1}{2}) (3 a - \frac{1}{2})$ .

$9 a^{2} - 2 a + \frac{1}{4}$

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$6 a^{2} - 3 a + \frac{1}{4}$

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$9 a^{2} - 3 a + \frac{1}{4}$

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$9 a^{2} - 3 a + \frac{1}{2}$

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Solution

The correct option is C
$9 a^{2} - 3 a + \frac{1}{4}$

$(3 a - \frac{1}{2}) (3 a - \frac{1}{2}) = (3 a - \frac{1}{2})^{2}$

Using the identity
$(a - b)^{2} = a^{2} - 2 a b + b^{2}$ ,

$(3 a - \frac{1}{2})^{2} = (3 a)^{2} - 2 \times 3 a \times \frac{1}{2} + (\frac{1}{2})^{2} = 9 a^{2} - 3 a + \frac{1}{4}$

Suggest Corrections

Find the value of (3a−12)(3a−12).

The correct option is C 9a2−3a+14 (3a−12)(3a−12)=(3a−12)2 Using the identity (a−b)2=a2−2ab+b2, (3a−12)2=(3a)2−2×3a×12+(12)2=9a2−3a+14

Find the value of $(3 a - \frac{1}{2}) (3 a - \frac{1}{2})$ .