NCERT Solutions for Class 9 Maths Chapter 4 — Exercise Set 4.2#
$$a^2 + 2ab + b^2 = (a + b)^2$$$$a^2 - 2ab + b^2 = (a - b)^2$$
To factor a perfect square: find $a$ and $b$ whose squares are the first and last terms, then check that the middle term is $2ab$. If the coefficients are awkward, first take out a common factor, as in Example 7.
1. Factor completely: #
(i) $9x^2 + 24xy + 16y^2$
$9x^2 = (3x)^2$ and $16y^2 = (4y)^2$. The middle term is $2(3x)(4y) = 24xy$ ✓. We factor $9x^2 + 24xy + 16y^2$:
$$= (3x)^2 + 2(3x)(4y) + (4y)^2$$$$= (3x + 4y)^2$$
(ii) $4s^2 + 20st + 25t^2$
$4s^2 = (2s)^2$ and $25t^2 = (5t)^2$. The middle term is $2(2s)(5t) = 20st$ ✓. We factor $4s^2 + 20st + 25t^2$:
$$= (2s)^2 + 2(2s)(5t) + (5t)^2$$$$= (2s + 5t)^2$$
(iii) $49x^2 + 28xy + 4y^2$
$49x^2 = (7x)^2$ and $4y^2 = (2y)^2$. The middle term is $2(7x)(2y) = 28xy$ ✓. We factor $49x^2 + 28xy + 4y^2$:
$$= (7x)^2 + 2(7x)(2y) + (2y)^2$$$$= (7x + 2y)^2$$
(iv) $64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2$
$64p^2 = (8p)^2$ and $\dfrac{4}{9}q^2 = \left(\dfrac{2}{3}q\right)^2$. The middle term is $2(8p)\left(\dfrac{2}{3}q\right) = \dfrac{32}{3}pq$ ✓. We factor $64p^2 + \dfrac{32}{3}pq + \dfrac{4}{9}q^2$:
$$= (8p)^2 + 2(8p)\left(\dfrac{2}{3}q\right) + \left(\dfrac{2}{3}q\right)^2$$$$= \left(8p + \dfrac{2}{3}q\right)^2$$
*(v) $3a^2 + 4ab + \dfrac{4}{3}b^2$
$3a^2$ is not the square of a simple term, so, as the hint suggests, we first take out the common factor 3. We factor $3a^2 + 4ab + \dfrac{4}{3}b^2$:
$$= 3\left(a^2 + \dfrac{4}{3}ab + \dfrac{4}{9}b^2\right)$$$$= 3\left(a^2 + 2(a)\left(\dfrac{2}{3}b\right) + \left(\dfrac{2}{3}b\right)^2\right)$$
$$= 3\left(a + \dfrac{2}{3}b\right)^2$$
(Taking out $\dfrac{1}{3}$ instead gives the same answer in another form: $\dfrac{1}{3}(9a^2 + 12ab + 4b^2) = \dfrac{1}{3}(3a + 2b)^2$.)
*(vi) $\dfrac{9}{5}s^2 + 6sv + 5v^2$
We take out the common factor $\dfrac{1}{5}$ so that the first and last terms become perfect squares. We factor $\dfrac{9}{5}s^2 + 6sv + 5v^2$:
$$= \dfrac{1}{5}\left(9s^2 + 30sv + 25v^2\right)$$$$= \dfrac{1}{5}\left((3s)^2 + 2(3s)(5v) + (5v)^2\right)$$
$$= \dfrac{1}{5}(3s + 5v)^2$$
2. Find the values of the following using the identity $(a - b)^2 = a^2 - 2ab + b^2$. #
(i) $(79)^2$
We write 79 as $80 - 1$. We find $(79)^2$:
$$= (80 - 1)^2$$$$= 80^2 - 2(80)(1) + 1^2$$
$$= 6400 - 160 + 1$$
$$= 6241$$
(ii) $(193)^2$
We write 193 as $200 - 7$. We find $(193)^2$:
$$= (200 - 7)^2$$$$= 200^2 - 2(200)(7) + 7^2$$
$$= 40000 - 2800 + 49$$
$$= 37249$$
(iii) $(299)^2$
We write 299 as $300 - 1$. We find $(299)^2$:
$$= (300 - 1)^2$$$$= 300^2 - 2(300)(1) + 1^2$$
$$= 90000 - 600 + 1$$
$$= 89401$$