Using the identity (a + b)² = a² + 2ab + b², expand the following: Class 9

Using the identity (a + b)² = a² + 2ab + b², expand the following: Class 9

Question 1.

Using the identity (a + b)² = a² + 2ab + b², expand the following: Class 9

i. (7x + 4y)²

ii. $\left(\frac{7}{5} x+\frac{3}{2} y\right)^2$

iii. (2.5p + 1.5q)²

iv. $\left(\frac{3}{4} s+8 t\right)^2$

v. $\left(x+\frac{1}{2 y}\right)^2$

vi. $\left(\frac{1}{x}+\frac{1}{y}\right)^2$

Solution:

i. Here, a = 7x and b = 4y

∴ (7x + 4y)² = (7x)² + 2(7x)(4y) + (4y)²

= 49x² + 56xy + 16y²

ii. Here, a = $\frac{7}{5}$x and b = $\frac{3}{2}$y

$\begin{aligned} \therefore \quad\left(\frac{7}{5} x+\frac{3}{2} y\right)^2 & =\left(\frac{7}{5} x\right)^2+2\left(\frac{7}{5} x\right)\left(\frac{3}{2} y\right)+\left(\frac{3}{2} y\right)^2 \\ & =\frac{49}{25} x^2+\frac{21}{5} x y+\frac{9}{4} y^2\end{aligned}$

iii. Here, a = 2.5p and b = 1.5q

∴ (2.5 p + 1.5q)² = (2.5 p)² + 2(2.5p)(1.5q) + (1.5q)²

= 6.15p² + 7.5pq + 2.25q²

iv. Here, a = $\frac{3}{4}$s and b = 8t

$\begin{aligned} \therefore \quad\left(\frac{3}{4} s+8 t\right)^2 & =\left(\frac{3}{4} s\right)^2+2\left(\frac{3}{4} s\right)(8 t)+(8 t)^2 \\ & =\frac{9}{16} s^2+12 s t+64 t^2\end{aligned}$

v. Here, a = x and b = $\frac{1}{2}$ y

$\begin{aligned} \therefore \quad\left(x+\frac{1}{2 y}\right)^2 & =x^2+2(x)\left(\frac{1}{2 y}\right)+\left(\frac{1}{2 y}\right)^2 \\ & =x^2+\frac{x}{y}+\frac{1}{4 y^2}\end{aligned}$

vi. Here, a = $\frac{1}{x}$ and b = $\frac{1}{y}$

$\begin{aligned} \therefore \quad\left(\frac{1}{x}+\frac{1}{y}\right)^2 & =\left(\frac{1}{x}\right)^2+2\left(\frac{1}{x}\right)\left(\frac{1}{y}\right)+\left(\frac{1}{y}\right)^2 \\ & =\frac{1}{x^2}+\frac{2}{x y}+\frac{1}{y^2}\end{aligned}$