The value of (12a−14b+2)2 is

(12a−14b+2)2 =(12a)2+(−14b)2+22+2⋅12⋅a⋅(−14b)+2(−14b)⋅2+2⋅12a⋅2 =14a2+116b2+4−ab4−b+2a =a24+b216+4−ab4−b+2a

The value of (12a−14b+2)2 is

(12a−14b+2)2 =(12a)2+(−14b)2+22+2⋅12⋅a⋅(−14b)+2(−14b)⋅2+2⋅12a⋅2 =14a2+116b2+4−ab4−b+2a =a24+b216+4−ab4−b+2a

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The value of ${(\frac{1}{2} a - \frac{1}{4} b + 2)}^{2}$ is

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\frac{a^{2}}{4} + \frac{b^{2}}{16} + 4 + \frac{a b}{4} - b + a

\frac{a^{2}}{4} + \frac{b^{2}}{16} + 4 - \frac{a b}{4} - b + 2 a

\frac{a^{2}}{4} - \frac{b^{2}}{16} + 4 + \frac{a b}{4} + b + 2 a

\frac{a^{2}}{4} + \frac{b^{2}}{16} + 2 - \frac{a b}{4} - b + 2 a

answer is B.

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Detailed Solution

{(\frac{1}{2} a - \frac{1}{4} b + 2)}^{2}

$= {(\frac{1}{2} a)}^{2} + {(\frac{- 1}{4} b)}^{2} + 2^{2} + 2 \cdot \frac{1}{2} \cdot a \cdot (- \frac{1}{4} b) + 2 (- \frac{1}{4} b) \cdot 2 + 2 \cdot \frac{1}{2} a \cdot 2$

$= \frac{1}{4} a^{2} + \frac{1}{16} b^{2} + 4 - \frac{a b}{4} - b + 2 a$

$= \frac{a^{2}}{4} + \frac{b^{2}}{16} + 4 - \frac{a b}{4} - b + 2 a$

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The value of ${(\frac{1}{2} a - \frac{1}{4} b + 2)}^{2}$ is