NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.3#

117 splits into three easy parts, $100 + 10 + 7$, so we use the $(a + b + c)^2$ identity. We find $117^2$:

$$= (100 + 10 + 7)^2$$

$$= 100^2 + 10^2 + 7^2 + 2(100)(10) + 2(10)(7) + 2(7)(100)$$

$$= 10000 + 100 + 49 + 2000 + 140 + 1400$$

$$= 13689$$

78 is close to 80, so we use $(a - b)^2$. We find $78^2$:

$$= (80 - 2)^2$$

$$= 80^2 - 2(80)(2) + 2^2$$

$$= 6400 - 320 + 4$$

$$= 6084$$

198 is close to 200, so we use $(a - b)^2$. We find $198^2$:

$$= (200 - 2)^2$$

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

$$= 40000 - 800 + 4$$

$$= 39204$$

214 splits into $200 + 10 + 4$, so we use $(a + b + c)^2$. We find $214^2$:

$$= (200 + 10 + 4)^2$$

$$= 200^2 + 10^2 + 4^2 + 2(200)(10) + 2(10)(4) + 2(4)(200)$$

$$= 40000 + 100 + 16 + 4000 + 80 + 1600$$

$$= 45796$$

1104 is $1100 + 4$, so we use $(a + b)^2$. We find $1104^2$:

$$= (1100 + 4)^2$$

$$= 1100^2 + 2(1100)(4) + 4^2$$

$$= 1210000 + 8800 + 16$$

$$= 1218816$$

1120 is $1100 + 20$, so we use $(a + b)^2$. We find $1120^2$:

$$= (1100 + 20)^2$$

$$= 1100^2 + 2(1100)(20) + 20^2$$

$$= 1210000 + 44000 + 400$$

$$= 1254400$$

$16y^2 = (4y)^2$ and $9 = 3^2$, and the middle term is $-2(4y)(3) = -24y$, so we use $(a - b)^2$. We factor $16y^2 - 24y + 9$:

$$= (4y)^2 - 2(4y)(3) + 3^2$$

$$= (4y - 3)^2$$

$\dfrac{9}{4}s^2 = \left(\dfrac{3}{2}s\right)^2$ and $4t^2 = (2t)^2$, and $2\left(\dfrac{3}{2}s\right)(2t) = 6st$, so we use $(a + b)^2$. We factor $\dfrac{9}{4}s^2 + 6st + 4t^2$:

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

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

There are three square terms, $\dfrac{m^2}{9} = \left(\dfrac{m}{3}\right)^2$, $\dfrac{k^2}{4} = \left(\dfrac{k}{2}\right)^2$ and $9n^2 = (3n)^2$, so we try $(a + b + c)^2$ with $a = \dfrac{m}{3}$, $b = \dfrac{k}{2}$, $c = 3n$. The other terms match:

$$2ab = 2\left(\dfrac{m}{3}\right)\left(\dfrac{k}{2}\right) = \dfrac{mk}{3}$$

$$2bc = 2\left(\dfrac{k}{2}\right)(3n) = 3nk$$

$$2ca = 2(3n)\left(\dfrac{m}{3}\right) = 2mn$$

So the expression factors as:

$$\dfrac{m^2}{9} + \dfrac{mk}{3} + \dfrac{k^2}{4} + 3nk + 2mn + 9n^2 = \left(\dfrac{m}{3} + \dfrac{k}{2} + 3n\right)^2$$

$\dfrac{p^2}{16} = \left(\dfrac{p}{4}\right)^2$ and $\dfrac{16}{p^2} = \left(\dfrac{4}{p}\right)^2$, and $2\left(\dfrac{p}{4}\right)\left(\dfrac{4}{p}\right) = 2$, so we use $(a - b)^2$. We factor $\dfrac{p^2}{16} - 2 + \dfrac{16}{p^2}$:

$$= \left(\dfrac{p}{4}\right)^2 - 2\left(\dfrac{p}{4}\right)\left(\dfrac{4}{p}\right) + \left(\dfrac{4}{p}\right)^2$$

$$= \left(\dfrac{p}{4} - \dfrac{4}{p}\right)^2$$

The squares are $(3a)^2$, $(2b)^2$ and $c^2$. The terms with $b$ and another letter are negative ($-12ab$, $-4bc$), and the $ac$ term is positive, so $b$ has the minus sign: we try $a \to 3a$, $b \to -2b$, $c \to c$. The other terms match:

$$2(3a)(-2b) = -12ab$$

$$2(-2b)(c) = -4bc$$

$$2(c)(3a) = 6ac$$

So the expression factors as:

$$9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bc = (3a - 2b + c)^2$$

Here $a = p$, $b = 3q$, $c = 7r$. We expand $(p + 3q + 7r)^2$:

$$= p^2 + (3q)^2 + (7r)^2 + 2(p)(3q) + 2(3q)(7r) + 2(7r)(p)$$

$$= p^2 + 9q^2 + 49r^2 + 6pq + 42qr + 14pr$$

Here $a = 3x$, $b = -2y$, $c = 4z$. We expand $(3x - 2y + 4z)^2$:

$$= (3x)^2 + (-2y)^2 + (4z)^2 + 2(3x)(-2y) + 2(-2y)(4z) + 2(4z)(3x)$$

$$= 9x^2 + 4y^2 + 16z^2 - 12xy - 16yz + 24xz$$

We expand each square with the $(a + b + c)^2$ identity:

$$(a + b - c)^2 = a^2 + b^2 + c^2 + 2ab - 2bc - 2ca$$

$$(a - b + c)^2 = a^2 + b^2 + c^2 - 2ab - 2bc + 2ca$$

$$(a - b - c)^2 = a^2 + b^2 + c^2 - 2ab + 2bc - 2ca$$

Adding the three expansions:

$$\begin{array}{rcl} (a + b - c)^2 &=& a^2 + b^2 + c^2 + 2ab - 2bc - 2ca \\ (a - b + c)^2 &=& a^2 + b^2 + c^2 - 2ab - 2bc + 2ca \\ (a - b - c)^2 &=& a^2 + b^2 + c^2 - 2ab + 2bc - 2ca \\ \hline \text{LHS} &=& 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ca \end{array}$$

The right-hand side is $2a^2 + 2b^2 + 2c^2$:

$$\Rightarrow 3a^2 + 3b^2 + 3c^2 - 2ab - 2bc - 2ca \neq 2a^2 + 2b^2 + 2c^2$$

For example, $a = b = c = 1$ gives $\text{LHS} = 1 + 1 + 1 = 3$ but $\text{RHS} = 6$.

∴ No, it is not an identity. It is true only for some values (when $a^2 + b^2 + c^2 = 2ab + 2bc + 2ca$), not for all values.