Question

$\huge{\mathfrak{{☆Question ☆}}}$

What is the perfect square form of 16x² - 4x + $\sf{\frac{9}{36}}$? ​

Answer 1

Answer :-

$\implies\sf 16x^2 - 4x + \frac{9}{36}$

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$\implies\sf (4x)^2 - 4x + \Big(\frac{3}{6}\Big)^2$

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$\implies\sf (4x)^2 - 2 ( 4x ) \Big( \dfrac{3}{6} \Big) + \Big(\dfrac{3}{6}\Big)^2$

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Using the identity

• $\sf (a-b)^2 = a^2 - 2ab - b^2$

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According to the expression, we got -

• $\sf a = 4x$
• $\sf b = \dfrac{3}{6}$

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$\implies\sf \Big(4x - \dfrac{3}{6}\Big)^2$

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$\boxed{\sf 16x^2 - 4x + \frac{9}{36} = \Big(4x - \dfrac{3}{6}\Big)^2 }$

Verification -

$\sf \Big(4x - \dfrac{3}{6}\Big)^2$

$\sf = (4x)^2 + \Big(\dfrac{3}{6}\Big)^2 - 2 \times 4x \times \Big(\dfrac{3}{6}\Big)$

$\sf = 16x^2 + \frac{9}{36} - 4x$

Hence verified.

Answer 2

$\huge { \bf{\underline{Answer :-}}}$

$\implies\sf 16x^2 - 4x + \frac{9}{36}$

$\implies\sf (4x)^2 - 4x + \Big(\frac{3}{6}\Big)^2$

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$\implies\sf (4x)^2 - 2 ( 4x ) \Big( \dfrac{3}{6} \Big) + \Big(\dfrac{3}{6}\Big)^2$

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$\large \blue { \boldsymbol{\underline{Using \: \: the \: \: identity }}}:$

$\sf (a-b)^2 = a^2 - 2ab - b^2(a−b)2=a2−2ab−b2$

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$\large \blue { \boldsymbol{\underline{According \: to \: the \: expression, \: : -}}}$

$\sf a = 4xa=4x$

$\sf b = \dfrac{3}{6}b=63$

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$\implies\sf \Big(4x - \dfrac{3}{6}\Big)^2$

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$\boxed{\sf 16x^2 - 4x + \frac{9}{36} = \Big(4x - \dfrac{3}{6}\Big)^2 }$

$\huge \bf{ \underline{Verification : -}}$

$\sf \Big(4x - \dfrac{3}{6}\Big)^2(4x−63)$

$\sf = (4x)^2 + \Big(\dfrac{3}{6}\Big)^2 - 2 \times 4x \times \Big(\dfrac{3}{6}\Big)$

$\sf = 16x^2 + \frac{9}{36} - 4x$

$\bigg \lgroup { \bf {\red {Hence \: \: verified.}}}\bigg \rgroup$

Hope it helps ♥️♥️♪••