Question

solve (2x + 3/4 y )( 2x - 1/2 y) by the identity (x+a)(x-b)​

Answer 1

Given :

$\sf \implies \bigg(2x + \dfrac{3}{4}y\bigg)\bigg(2x - \dfrac{1}{2}y\bigg)$

Identity used :

$\large\implies\boxed{ \boxed{ \sf(x + a)(x + b) = {x}^{2} + (a + b)x + ab}}$

Solution :

$\:\:\:\bullet \:\:\: \sf x = 2x \\ \\\:\:\: \bullet\:\:\: \sf a =\dfrac{3}{4}y \\ \\\:\:\:\bullet \sf \:\:\:b =- \dfrac{1}{2}y\\ \\ \implies\sf{(2x)}^{2} + \bigg(\frac{3}{4} y - \frac{1}{2} y \bigg)2x + \frac{3}{4}y \times - \bigg(\frac{1}{2}y \bigg) \\ \\ \sf \implies {4x}^{2} + \bigg( \frac{3y - 2y}{4}\bigg)2x - \frac{3}{8}{y}^{2} \\ \\\implies \sf 4 {x}^{2} + \bigg(\frac{1}{4} y\bigg)2x - \frac{3}{8} {y}^{2} \\ \\ \implies \sf4 {x}^{2}+ \frac{xy}{2} - \frac{3}{8}{y}^{2}$

Answer 2

AnswEr:-

$\sf \bigg(2x + \dfrac{3}{4}y \bigg) \bigg(2x - \dfrac{1}{2}y \bigg)$

$\sf \implies x = 2x$

$\sf \implies a = \dfrac{3}{4}y$

$\sf \implies b = \dfrac{-1}{2}y$

★ Identity used:-

$\large\bold{\underline{\underline{\boxed{\sf{\purple{(x + a)(x + b) = x^2 + (a + b)x + ab}}}}}}$

★ Putting values in indentity:-

• $\sf x^2 + (a + b)x + ab$

• $\sf (2x)^2 + \bigg( \dfrac{3}{4}y - \dfrac{1}{2}y \bigg)2x + \dfrac{3}{4}y × - \dfrac{1}{2}y$

• $\sf 4x^2 + \bigg( \dfrac{3y - 2y}{4} \bigg)2x - \dfrac{-3}{8}y^2$

• $\sf 4x^2 + \bigg( \dfrac{1}{4}y \bigg)2x - \dfrac{3}{8}y^2$

• $\sf 4x^2 + \dfrac{xy}{2} - \dfrac{3}{8}y^2$

★ Some related formula for quadratic equations are:-

• $\sf (x + y)^2 = x^2 + y^2 + 2xy$

• $\sf (x - y)^2 = x^2 + y^2 - 2xy$

• $\sf (x + y)(x - y) = x^2 - y^2$

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