Mathematics solution NCERT
Class 9 - Chapter 4: Exploring Algebraic Identities
Question 1. Factor completely:
(i) $9x^2+24xy+16y^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Compare the given expression with the identity.
$$ 9x^2+24xy+16y^2 $$Here,
$$ 9x^2=(3x)^2, $$ $$ 16y^2=(4y)^2, $$and
$$ 2(3x)(4y)=24xy. $$Thus, the given expression is of the form $a^2+2ab+b^2$, where
$$ a=3x \qquad\text{and}\qquad b=4y. $$Using the identity, we get
$$ \begin{aligned} 9x^2+24xy+16y^2 &=(3x+4y)^2. \end{aligned} $$Answer:
$$ \boxed{(3x+4y)^2} $$(ii) $4s^2+20st+25t^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Compare the given expression with the identity.
$$ 4s^2+20st+25t^2 $$Here,
$$ 4s^2=(2s)^2, $$ $$ 25t^2=(5t)^2, $$and
$$ 2(2s)(5t)=20st. $$Thus, the given expression is of the form $a^2+2ab+b^2$, where
$$ a=2s \qquad\text{and}\qquad b=5t. $$Using the identity, we obtain
$$ \begin{aligned} 4s^2+20st+25t^2 &=(2s+5t)^2. \end{aligned} $$Answer:
$$ \boxed{(2s+5t)^2} $$(iii) $49x^2+28xy+4y^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Compare the given expression with the identity.
$$ 49x^2+28xy+4y^2 $$Here,
$$ 49x^2=(7x)^2, $$ $$ 4y^2=(2y)^2, $$and
$$ 2(7x)(2y)=28xy. $$Thus, the given expression is of the form $a^2+2ab+b^2$, where
$$ a=7x \qquad\text{and}\qquad b=2y. $$Using the identity, we get
$$ \begin{aligned} 49x^2+28xy+4y^2 &=(7x+2y)^2. \end{aligned} $$Answer:
$$ \boxed{(7x+2y)^2} $$Question 1. Factor completely:
(iv) $64p^2+\dfrac{32}{3}pq+\dfrac{4}{9}q^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$Compare the given expression with the identity.
$$ 64p^2+\frac{32}{3}pq+\frac49q^2 $$Here,
$$ 64p^2=(8p)^2, $$ $$ \frac49q^2=\left(\frac23q\right)^2, $$and
$$ 2(8p)\left(\frac23q\right)=\frac{32}{3}pq. $$Thus, the given expression is of the form $a^2+2ab+b^2$, where
$$ a=8p \qquad\text{and}\qquad b=\frac23q. $$Using the identity, we get
$$ \begin{aligned} 64p^2+\frac{32}{3}pq+\frac49q^2 &=\left(8p+\frac23q\right)^2. \end{aligned} $$Answer:
$$ \boxed{\left(8p+\frac23q\right)^2} $$(v) $3a^2+4ab+\dfrac43b^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$First, write each perfect square.
$$ 3a^2=(\sqrt3\,a)^2, $$ $$ \frac43b^2=\left(\frac{2}{\sqrt3}b\right)^2. $$Now find twice the product of these terms.
$$ \begin{aligned} 2(\sqrt3\,a)\left(\frac{2}{\sqrt3}b\right) &=4ab. \end{aligned} $$This is exactly the middle term of the given expression.
Therefore, the given expression is of the form $a^2+2ab+b^2$, where
$$ a=\sqrt3\,a \qquad\text{and}\qquad b=\frac{2}{\sqrt3}b. $$Using the identity, we obtain
$$ \begin{aligned} 3a^2+4ab+\frac43b^2 &=\left(\sqrt3\,a+\frac{2}{\sqrt3}b\right)^2. \end{aligned} $$Answer:
$$ \boxed{\left(\sqrt3\,a+\frac{2}{\sqrt3}b\right)^2} $$(vi) $\dfrac95s^2+6sv+5v^2$
Solution:
We know that
$$ (a+b)^2=a^2+2ab+b^2. $$First, write each perfect square.
$$ \frac95s^2=\left(\frac3{\sqrt5}s\right)^2, $$ $$ 5v^2=(\sqrt5\,v)^2. $$Now find twice the product of these terms.
$$ \begin{aligned} 2\left(\frac3{\sqrt5}s\right)(\sqrt5\,v) &=6sv. \end{aligned} $$This is exactly the middle term of the given expression.
Hence, the given expression is of the form $a^2+2ab+b^2$, where
$$ a=\frac3{\sqrt5}s \qquad\text{and}\qquad b=\sqrt5\,v. $$Using the identity, we get
$$ \begin{aligned} \frac95s^2+6sv+5v^2 &=\left(\frac3{\sqrt5}s+\sqrt5\,v\right)^2. \end{aligned} $$Answer:
$$ \boxed{\left(\frac3{\sqrt5}s+\sqrt5\,v\right)^2} $$Question 2. Find the values of the following using the identity $(a-b)^2=a^2-2ab+b^2$.
(i) $(79)^2$
Solution:
We know that
$$ (a-b)^2=a^2-2ab+b^2. $$Express $79$ as the difference of two convenient numbers.
$$ 79=80-1. $$Here,
$$ a=80 \qquad\text{and}\qquad b=1. $$Using the identity,
$$ \begin{aligned} 79^2 &=(80-1)^2\\ &=80^2-2(80)(1)+1^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 80^2&=6400,\\[4pt] 2(80)(1)&=160,\\[4pt] 1^2&=1. \end{aligned} $$Substitute these values.
$$ \begin{aligned} 79^2 &=6400-160+1\\ &=6241. \end{aligned} $$Answer:
$$ \boxed{79^2=6241} $$(ii) $(193)^2$
Solution:
Express $193$ as
$$ 193=200-7. $$Here,
$$ a=200 \qquad\text{and}\qquad b=7. $$Using the identity,
$$ \begin{aligned} 193^2 &=(200-7)^2\\ &=200^2-2(200)(7)+7^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 200^2&=40000,\\[4pt] 2(200)(7)&=2800,\\[4pt] 7^2&=49. \end{aligned} $$Substitute these values.
$$ \begin{aligned} 193^2 &=40000-2800+49\\ &=37249. \end{aligned} $$Answer:
$$ \boxed{193^2=37249} $$(iii) $(299)^2$
Solution:
Express $299$ as
$$ 299=300-1. $$Here,
$$ a=300 \qquad\text{and}\qquad b=1. $$Using the identity,
$$ \begin{aligned} 299^2 &=(300-1)^2\\ &=300^2-2(300)(1)+1^2. \end{aligned} $$Now calculate each term.
$$ \begin{aligned} 300^2&=90000,\\[4pt] 2(300)(1)&=600,\\[4pt] 1^2&=1. \end{aligned} $$Substitute these values.
$$ \begin{aligned} 299^2 &=90000-600+1\\ &=89401. \end{aligned} $$Answer:
$$ \boxed{299^2=89401} $$