NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.1#

$$(a + b)^2 = a^2 + 2ab + b^2$$

To use the identity, decide what $a$ and $b$ are, then write $a^2$, $2ab$ and $b^2$ one after another.

1. Using the identity $(a + b)^2 = a^2 + 2ab + b^2$, expand the following: #

(i) $(7x + 4y)^2$

Here $a = 7x$ and $b = 4y$. We expand $(7x + 4y)^2$:

$$= (7x)^2 + 2(7x)(4y) + (4y)^2$$

$$= 49x^2 + 56xy + 16y^2$$
(ii) $\left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2$

Here $a = \dfrac{7}{5}x$ and $b = \dfrac{3}{2}y$. We expand $\left(\dfrac{7}{5}x + \dfrac{3}{2}y\right)^2$:

$$= \left(\dfrac{7}{5}x\right)^2 + 2\left(\dfrac{7}{5}x\right)\left(\dfrac{3}{2}y\right) + \left(\dfrac{3}{2}y\right)^2$$

$$= \dfrac{49}{25}x^2 + \dfrac{21}{5}xy + \dfrac{9}{4}y^2$$
(iii) $(2.5p + 1.5q)^2$

Here $a = 2.5p$ and $b = 1.5q$. We expand $(2.5p + 1.5q)^2$:

$$= (2.5p)^2 + 2(2.5p)(1.5q) + (1.5q)^2$$

$$= 6.25p^2 + 7.5pq + 2.25q^2$$
(iv) $\left(\dfrac{3}{4}s + 8t\right)^2$

Here $a = \dfrac{3}{4}s$ and $b = 8t$. We expand $\left(\dfrac{3}{4}s + 8t\right)^2$:

$$= \left(\dfrac{3}{4}s\right)^2 + 2\left(\dfrac{3}{4}s\right)(8t) + (8t)^2$$

$$= \dfrac{9}{16}s^2 + 12st + 64t^2$$
(v) $\left(x + \dfrac{1}{2y}\right)^2$

Here $a = x$ and $b = \dfrac{1}{2y}$. We expand $\left(x + \dfrac{1}{2y}\right)^2$:

$$= x^2 + 2(x)\left(\dfrac{1}{2y}\right) + \left(\dfrac{1}{2y}\right)^2$$

$$= x^2 + \dfrac{x}{y} + \dfrac{1}{4y^2}$$
(vi) $\left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2$

Here $a = \dfrac{1}{x}$ and $b = \dfrac{1}{y}$. We expand $\left(\dfrac{1}{x} + \dfrac{1}{y}\right)^2$:

$$= \left(\dfrac{1}{x}\right)^2 + 2\left(\dfrac{1}{x}\right)\left(\dfrac{1}{y}\right) + \left(\dfrac{1}{y}\right)^2$$

$$= \dfrac{1}{x^2} + \dfrac{2}{xy} + \dfrac{1}{y^2}$$

2. Using the same identity, find the values of the following: #

(i) $(64)^2$

We write 64 as $60 + 4$, so $a = 60$ and $b = 4$. We find $(64)^2$:

$$= (60 + 4)^2$$

$$= 60^2 + 2(60)(4) + 4^2$$

$$= 3600 + 480 + 16$$

$$= 4096$$
(ii) $(105)^2$

We write 105 as $100 + 5$. We find $(105)^2$:

$$= (100 + 5)^2$$

$$= 100^2 + 2(100)(5) + 5^2$$

$$= 10000 + 1000 + 25$$

$$= 11025$$
(iii) $(205)^2$

We write 205 as $200 + 5$. We find $(205)^2$:

$$= (200 + 5)^2$$

$$= 200^2 + 2(200)(5) + 5^2$$

$$= 40000 + 2000 + 25$$

$$= 42025$$

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