Exercise 3.1 - Chapter 3 Algebra Term 3 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $3.1$
Question $1 .$
Fill in the blanks.
1. $(\mathrm{p}-\mathrm{q})^{2}=$
2. The product of $(x+5)$ and $(x-5)$ is
3. The factors of $x^{2}-4 x+4$ are
4. Express $24 \mathrm{ab}^{2} \mathrm{c}^{2}$ as product of its factors is
Answers:
1. $\mathrm{p}^{2}-2 \mathrm{pq}+\mathrm{q}^{2}$
2. $x^{2}-25$
3. $(x-2)$ and $(x-2)$
4. $2 \times 2 \times 2 \times 3 \times a \times b \times b \times c \times c$
 

Question $2 .$
Say whether the following statements are True or False.
(i) $(7 x+3)(7 x-4)=49 x^{2}-7 x-12$
(ii) $(a-1)^{2}=a^{2}-1$.
(iii) $\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)\left(\mathrm{y}^{2}+\mathrm{x}^{2}\right)=\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)^{2}$
(iv) $2 p$ is the factor of $8 \mathrm{pq}$.
Answers:
(i) True
(ii) False
(iii) True
(iv) True

Question $3 .$
Express the following as the product of its factors.
(i) $24 a b^{2} c^{2}$
(ii) $36 \mathrm{x}^{3} \mathrm{y}^{2} \mathrm{z}$
(iii) $56 \mathrm{mn}^{2} \mathrm{p}^{2}$
Solution:
(i) $24 \mathrm{ab}^{2} \mathrm{c}^{2}=2 \times 2 \times 2 \times 3 \times \mathrm{a} \times \mathrm{b} \times \mathrm{b} \times \mathrm{c} \times \mathrm{c}$
(ii) $36 \mathrm{x}^{3} \mathrm{y}^{2} \mathrm{z}=2 \times 2 \times 3 \times 3 \times \mathrm{x} \times \mathrm{x} \times \mathrm{x} \times \mathrm{y} \times \mathrm{y} \times \mathrm{z}$
(iii) $56 \mathrm{mn}^{2} \mathrm{p}^{2}=2 \times 2 \times 2 \times 7 \times \mathrm{m} \times \mathrm{n} \times \mathrm{n} \times \mathrm{p} \times \mathrm{p}$
 

Question $4 .$
Using the identity $(x+a)(x+b)-x^{2}+x(a+b)+a b$, find the following product.
(i) $(x+3)(x+7)$
(ii) $(6 a+9)(6 a-5)$
(iii) $(4 \mathrm{x}+3 \mathrm{y})(4 \mathrm{x}+5 \mathrm{y})$
(iv) $(8+\mathrm{pq})(\mathrm{pq}+7)$
Solution:
(i) $(x+3)(x+7)$
Let $a=3 ; b=7$, then
$(x+3)(x+7)$ is of the form $x^{2}+x(a+b)+a b$
$(x+3)(x+7)=x^{2}+x(3+7)+(3 \times 7)=x^{2}+10 x+21$
(ii) $(6 a+9)(6 a-5)$
Substituting $x=6 a ; a=9$ and $b=-5$
In $(x+a)(x+b)=x^{2}+x(a+b)+a b$, we get
$(6 a+9)(6 a-5)=(6 a)^{2}+6 a(9+(-5))+(9 \times(-5))$
$6^{2} a^{2}+6 a(4)+(-45)=36 a^{2}+24 a-45$
$(6 a+9)(6 a-5)=36 a^{2}+24 a-45$

(iii) $(4 x+3 y)(4 x+5 y)$
Substituting $x=4 x ; a=3 y$ and $b=5 y$ in
$(x+a)(x+b)=x^{2}+x(a+b)+a b$, we get
$(4 x+3 y)(4 x-5 y)=(4 x)^{2}+4 x(3 y+5 y)+(3 y)(5 y)$
$=4^{2} x^{2}+4 x(8 y)+15 y^{2}=16 x^{2}+32 x y+15 y^{2}$
$(4 x+3 y)(4 x+5 y)=16 x^{2}+32 x y+15 y^{2}$
(iv) $(8+\mathrm{pq})(\mathrm{pq}+7)$
Substituting $\mathrm{x}=\mathrm{pq} ; \mathrm{a}=8$ and $\mathrm{b}=7$ in
$(x+a)(x+b)=x^{2}+x(a+b)+a b$, we get
$(\mathrm{pq}+8)(\mathrm{pq}+7)=(\mathrm{pq})^{2}+\mathrm{pq}(8+7)+(8)(7)$
$=\mathrm{p}^{2} \mathrm{q}^{2}+\mathrm{pq}(15)+56$
$(8+\mathrm{pq})(\mathrm{pq}+7)=\mathrm{p}^{2} \mathrm{q}^{2}+15 \mathrm{pq}+56$
 

Question $5 .$
Expand the following squares, using suitable identities.
(i) $(2 x+5)^{2}$
(ii) $(b-7)^{2}$
(iii) $(\mathrm{mn}+3 \mathrm{p})^{2}$
(iv) $(x y z-1)^{2}$
Solution:

(i) $(2 x+5)^{2}$
Comparing $(2 x+5)^{2}$ with $(a+b)^{2}$ we have $a=2 x$ and $b=5$ $\mathrm{a}=2 \mathrm{x}$ and $\mathrm{b}=5$,
$(a+b)^{2}=a^{2}+2 a b+b^{2}$ $(2 x+5)^{2}=(2 x)^{2}+2(2 x)(5)+5^{2}=2^{2} x^{2}+20 x+25$ $=2^{2} x^{2}+20 x+25$ $(2 x+5)^{2}=4 x^{2}+20 x+25$
$\begin{aligned}
&(2 x+5)^{2}=(2 x)^{2}+2(2 x)(5)+5^{2}=2^{2} x^{2}+20 x+25 \\
&=2^{2} x^{2}+20 x+25 \\
&(2 x+5)^{2}=4 x^{2}+20 x+25
\end{aligned}$
(ii) $(b-7)^{2}$
Comparing $(b-7)^{2}$ with $(a-b)^{2}$ we have $a=b$ and $b=7$
$\begin{aligned}
&(a-b)^{2}=a^{2}-2 a b+b^{2} \\
&(b-7)^{2}=b^{2}-2(b)(7)+7^{2} \\
&(b-7)^{2}=b^{2}-14 b+49
\end{aligned}$
(iii) $(\mathrm{mn}+3 \mathrm{p})^{2}$
Comparing $(m n+3 p)^{2}$ with $(a+b)^{2}$ we have $(a+b)^{2}=a^{2}+2 a b+b^{2}$ $(m n+3 p)^{2}=(m n)^{2}+2(m n)(3 p)+(3 p)^{2}$ $(m n+3 p)^{2}=m^{2} n^{2}+6 m n p+9 p^{2}$
$\begin{aligned}
&(a+b)^{2}=a^{2}+2 a b+b^{2} \\
&(m n+3 p)^{2}=(m n)^{2}+2(m n)(3 p)+(3 p)^{2} \\
&(m n+3 p)^{2}=m^{2} n^{2}+6 m n p+9 p^{2}
\end{aligned}$
(iv) $(x y z-1)^{2}$
Comparing $(x y z-1)^{2}$ with $(a-b)^{2}$ we have $=a+x y z$ and $b=1$
$a=x y z$ and $b=1$
$(a-b)^{2}=a^{2}-2 a b+b^{2}$

$\begin{aligned}
&(x y z-1)^{2}=(x y z)^{2}-2(x y z)(1)+1^{2} \\
&(x y z-1)^{2}=x^{2} y^{2} z^{2}-2 x y z+1
\end{aligned}$

Question $6 .$
Using the identity $(a+b)(a-b)=a^{2}-b^{2}$, find the following product.
(i) $(\mathrm{p}+2)(\mathrm{p}-2)$
(ii) $(1+3 b)(3 b-1)$
(iii) $(4-\mathrm{mn})(\mathrm{mn}+4)$
(iv) $(6 x+7 y)(6 x-7 y)$
Solution:
(i) $(\mathrm{p}+2)(\mathrm{p}-2)$
Substituting $a=p ; b=2$ in the identity $(a+b)(a-b)=a^{2}-b^{2}$, we get $(p+2)(p-2)=p^{2}-2^{2}$
(ii) $(1+3 b)(3 b-1)$
$(1+3 b)(3 b-1)$ can be written as $(3 b+1)(3 b-1)$
Substituting $a=36$ and $b=1$ in the identity
$(a+b)(a-b)=a^{2}-b^{2}$, we get
$\begin{aligned}
&(3 b+1)(3 b-1)=(3 b)^{2}-1^{2}=3^{2} \times b^{2}-1^{2} \\
&(3 b+1)(3 b-1)=9 b^{2}-1^{2}
\end{aligned}$
(iii) $(4-\mathrm{mn})(\mathrm{mn}+4)$
$(4-\mathrm{mn})(\mathrm{mn}+4)$ can be written as $(4-\mathrm{mn})(4+\mathrm{mn})=(4+\mathrm{mn})(4-\mathrm{mn})$
Substituting $a=4$ and $b=m n$ is
$(a+b)(a-b)=a^{2}-b^{2}$, we get
$(4+m n)(4-m n)=4^{2}-(m n)^{2}=16-m^{2} n^{2}$
(iv) $(6 x+7 y)(6 x-7 y)$
Substituting $a=6 x$ and $b=7 y$ in
$(a+b)(a-b)=a^{2}-b^{2}$, We get
$(6 x+7 y)(6 x-7 y)=(6 x)^{2}-(7 y)^{2}=6^{2} x^{2}-7^{2} y^{2}$
$(6 x+7 y)(6 x-7 y)=(6 x)^{2}-(7 y)^{2}=6^{2} x^{2}-7^{2} y^{2}$
$(6 x+7 y)(6 x-7 y)=36 x^{2}-49 y^{2}$

Question $7 .$
Evaluate the following, using suitable identity.
(i) $51^{2}$
(ii) $103^{2}$
(iii) $998^{2}$
(iv) $47^{2}$
(v) $297 \times 303$
(vi) $990 \times 1010$
(vii) $51 \times 52$
Solution:
$51^{2}$
$=(50+1)^{2}$
Taking $a=50$ and $b=1$ we get
$(a+b)^{2}=a^{2}+2 a b+b^{2}$
$(50+1)^{2}=50^{2}+2(50)(1)+1^{2}=2500+100+1$
$51^{2}=2601$

(ii) $103^{2}$
$103^{2}=(100+3)^{2}$
Taking $a=100$ and $b=3$
$\begin{aligned}
&(a+b)^{2}=a^{2}+2 a b+b^{2} \text { becomes } \\
&(100+3)^{2}=100^{2}+2(100)(3)+3^{2}=10000+600+9 \\
&103^{2}=10609
\end{aligned}$
(iii) $998^{2}$
$998^{2}=(1000-2)^{2}$
Taking $a=1000$ and $b=2$
$\begin{aligned}
&(a-b)^{2}=a^{2}+2 a b+b^{2} \text { becomes } \\
&(1000-2)^{2}=1000^{2}-2(1000)(2)+2^{2} \\
&=1000000-4000+4 \\
&998^{2}=10,04,004
\end{aligned}$
(iv) $47^{2}$
$47^{2}=(50-3)^{2}$
Taking $a=50$ and $b=3$
$(a-b)^{2}=a^{2}-2 a b+b^{2}$ becomes
$\begin{aligned}
&(50-3)^{2}=50^{2}-2(50)(3)+3^{2} \\
&=2500-300+9=2200+9 \\
&47^{2}=2209
\end{aligned}$

(v) $297 \times 303$
$297 \times 303=(300-3)(300+3)$
Taking $a=300$ and $b=3$, then
$(a+b)(a-b)=a^{2}-b^{2}$ becomes
$(300+3)(300-3)=300^{2}-3^{2}$
$303 \times 297=90000-9$
$297 \times 303=89,991$
(vi) $990 \times 1010$
$990 \times 1010=(1000-10)(1000+10)$
Taking $\mathrm{a}=1000$ and $\mathrm{b}=10$, then
$(a-b)(a+b)=a^{2}-b^{2}$ becomes
$(1000-10)(1000+10)=1000^{2}-10^{2}$
$990 \times 1010=1000000-100$
$990 \times 1010=999900$
(vii) $51 \times 52$
$=(50+1)(50+1)$
Taking $\mathrm{x}=50, \mathrm{a}=1$ and $\mathrm{b}=2$
then $(x+a)(x+b)=x^{2}+(a+b) x+a b$ becomes
$(50+1)(50+2)=50^{2}+(1+2) 50+(1 \times 2)$
$2500+(3) 50+2=2500+150+2$
$51 \times 52=2652$
 

Question $8 .$
Simplify: $(a+b)^{2}-4 a b$
Solution:
$(a+b)^{2}-4 a b=a^{2}+b^{2}+2 a b-4 a b=a^{2}+b^{2}-2 a b=(a-b)^{2}$
 

Question $9 .$
Show that $(m-n)^{2}+(m+n)^{2}=2\left(m^{2}+n^{2}\right)$
Solution:

Taking the LHS $=(m-n)^{2}+(m+n)^{2}$
$\begin{aligned}
&=m^{2}-2 m n+n^{2}+m^{2}+2 m n+n^{2}=m^{2}+n^{2}+m^{2}+n^{2} \\
&=2 m^{2}+2 n^{2}
\end{aligned}$
$\text { RHS }$
$\left[\because(a+b)^{2}-4 a b=a^{2}+2 a b+b^{2}\right.$
$\begin{aligned} &=2\left(m^{2}+n^{2}\right)=\mathrm{RHS} \\ \therefore(m-n)^{2}+(m+n)^{2} &=2\left(m^{2}+n^{2}\right) \end{aligned}$
$\left.(a-b)^{2}=a^{2}-2 a b+b^{2}\right]$
$\therefore(m-n)^{2}+(m+n)^{2}=2\left(m^{2}+n^{2}\right)$

Question 10 .
If $\mathrm{a}+\mathrm{b}=10$, and $\mathrm{ab}=18$, find the value of $\mathrm{a}^{2}+\mathrm{b}^{2}$.
Solution:
We have $(a+b)^{2}=a^{2}+2 a b+b^{2}$
$(a+b)^{2}=a^{2}+b^{2}+2 a b$
given $a+b=0$ and $a b=18$
$\begin{aligned}
&10^{2}==a^{2}+b^{2}+2(18) \\
&100==a^{2}+b^{2}+36 \\
&100-36-a^{2}+b^{2} \\
&a^{2}+b^{2}=64
\end{aligned}$

Question $11 .$
Factorise the following algebraic expressions by using the identity $a^{2}-b^{2}=(a+b)(a-b)$.
(i) $\mathrm{z}^{2}-16$
(ii) $9-4 y^{2}$
(iii) $25 \mathrm{a}^{2}-49 \mathrm{~b}^{2}$
(iv) $x^{4}-y^{4}$
Solution:
(i) $\mathrm{z}^{2}-16$
$z^{2}-16=z^{2}-4^{2}$
We have $a^{2}-b^{2}=(a+b)(a-b)$
let $\mathrm{a}=\mathrm{z}$ and $\mathrm{b}=4$,
$z^{2}-4^{2}=(z+4)(z-4)$
(ii) $9-4 y^{2}$
$9-4 y^{2}=3^{2}-2^{2} y^{2}=3^{2}-(2 y)^{2}$
let $\mathrm{a}=3$ and $\mathrm{b}=2 \mathrm{y}$, then
$\begin{aligned}
&a^{2}-b^{2}=(a+b)(a-b) \\
&\therefore 3^{2}-(2 y)^{2}=(3+2 y)(3-2 y) \\
&9-4 y^{2}=(3+2 y)(3-2 y)
\end{aligned}$
(iii) $25 \mathrm{a}^{2}-49 \mathrm{~b}^{2}$ $25 \mathrm{a} 2-49 \mathrm{~b} 2=52-\mathrm{a} 2-72=(5 \mathrm{a}) 2-(7 \mathrm{~b}) 2$ let $\mathrm{A}=5 \mathrm{a}$ and $\mathrm{B}=7 \mathrm{~b}$ $\mathrm{~A}^{2} \mathrm{~B}^{2}$ $(5 \mathrm{a})^{2}-(7 \mathrm{~b})^{2}=(5 \mathrm{a}+7 \mathrm{~b})(5 \mathrm{a}-7 \mathrm{~b})$
$\begin{aligned}
&25 \mathrm{a} 2-49 \mathrm{~b} 2=52-\mathrm{a} 2-72=(5 \mathrm{a}) 2-(7 \mathrm{~b}) 2 \\
&\text { let } \mathrm{A}=5 \mathrm{a} \text { and } \mathrm{B}=7 \mathrm{~b} \\
&\mathrm{~A}^{2} \mathrm{~B}^{2} \\
&(5 \mathrm{a})^{2}-(7 \mathrm{~b})^{2}=(5 \mathrm{a}+7 \mathrm{~b})(5 \mathrm{a}-7 \mathrm{~b})
\end{aligned}$

(iv) $x^{4}-y^{4}$
Let $x^{4}-y^{4}=\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}$
We have $a^{2}-b^{2}=(a+b)(a-b)$
$\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}=\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)$
$x^{4}-y^{4}=\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)$
Again we have $x^{2}-y^{2}=(x+y)(x-y)$
$\therefore x^{4}-y^{4}=\left(x^{2}+y^{2}\right)(x+y)(x-y)$
 

Question $12 .$
Factorise the following using suitable identity.
(i) $x^{2}-8 x+16$
(ii) $y^{2}+20 y+100$
(iii) $36 \mathrm{~m}^{2}+60 \mathrm{~m}+25$
(iv) $64 \mathrm{x}^{2}-112 \mathrm{xy}+49 \mathrm{y}^{2}$
(v) $a^{2}+6 a b+9 b^{2}-c^{2}$
Solution:
(i) $x^{2}-8 x+16$
$x^{2}-8 x+16=x^{2}-(2 \times 4 \times x)+4^{2}$
This expression is in the form of identity
$\begin{aligned}
&a^{2}-2 a b+b^{2}=(a-b)^{2} \\
&x^{2}-2 \times 4 \times x+4^{2}=(x-4)^{2} \\
&\therefore x^{2}-8 x+16=(x-4)(x-4)
\end{aligned}$
(ii) $y^{2}+20 y+100$
$\begin{aligned}
&y^{2}+20 y+100=y^{2}+(2 \times(10)) y+(10 \times 10) \\
&=y^{2}+(2 \times 10 \times y)+10^{2}
\end{aligned}$
This is of the form of identity
$\begin{aligned}
&a^{2}+2 a b+b^{2}=(a+b)^{2} \\
&y^{2}+(2 \times 10 \times y)+10^{2}=(y+10)^{2} \\
&y^{2}+20 y+100=(y+10)^{2} \\
&y^{2}+20 y+100=(y+10)(y+10)
\end{aligned}$

(iii) $36 \mathrm{~m}^{2}+60 \mathrm{~m}+25$
$36 \mathrm{~m}^{2}+60 \mathrm{~m}+25=62 \mathrm{~m}^{2}+2 \times 6 \mathrm{~m} \times 5+5^{2}$
This expression is of the form of identity
$\begin{aligned}
&a^{2}+2 a b+b^{2}=\{a+b)^{2} \\
&(6 m)^{2}+(2 \times 6 m \times 5)+5^{2} \\
&=(6 m+5)^{2} \\
&36 m^{2}+60 m+25=(6 m+5)(6 m+5)
\end{aligned}$
(iv) $64 x^{2}-112 x y+49 y^{2}$
$64 x^{2}-112 x y+49 y^{2}=82 x^{2}-(2 \times 8 x \times 7 y)+7^{2} y^{2}$
This expression is of the form of identity
$\begin{aligned}
&a^{2}-2 a b+b^{2}=(a-b)^{2} \\
&(8 x)^{2}-(2 \times 8 x \times 7 y)+(7 y)^{2}=(8 x-7 y)^{2} \\
&64 x^{2}-112 x y+49 y^{2}=(8 x-7 y)(8 x-7 y)
\end{aligned}$
(v) $a^{2}+6 a b+9 b^{2}-c^{2}$ $a^{2}+6 a b+9 b^{2}-c^{2}=a^{2}+2 \times a \times 3 b+3^{2} b^{2}-c^{2}$ $=a^{2}+(2 \times a \times 3 b)+(3 b)^{2}-c^{2}$
$\begin{aligned}
&a^{2}+6 a b+9 b^{2}-c^{2}=a^{2}+2 \times a \times 3 b+3^{2} b^{2}-c^{2} \\
&=a^{2}+(2 \times a \times 3 b)+(3 b)^{2}-c^{2}
\end{aligned}$
This expression is of the form of identity
$\begin{aligned}
&{\left[a^{2}+2 a b+b^{2}\right]-c^{2}=(a+b)^{2}-c^{2}} \\
&a^{2}+(2 \times a \times 36)+(3 b)^{2}-c^{2}=(a+3 b)^{2}-c^{2}
\end{aligned}$
${\left[\mathrm{a}^{2}+2 \mathrm{ab}+\mathrm{b}^{2}\right]-\mathrm{c}^{2}=(\mathrm{a}+\mathrm{b})^{2}-\mathrm{c}^{2} }$ $\mathrm{a}^{2}+(2 \times \mathrm{a} \times 36)+(3 \mathrm{~b})^{2}-\mathrm{c}^{2}=(\mathrm{a}+3 \mathrm{~b})^{2}-\mathrm{c}^{2}$ Again this RHS is of the form of identity

Objective Type Questions
Question 1.
If $\mathrm{a}+\mathrm{b}=5$ and $\mathrm{a}^{2}+\mathrm{b}^{2}=13$, then $\mathrm{ab}=$ ?
(i) 12
(ii) 6
(iii) 5
(iv) 13
Answer:
(ii) 6
Hint: $(a+b)^{2}=25$
$13+2 \mathrm{ab}=25$
$2 a b=12$
$a b=6$
 

Question $2 .$
$(5+20)(-20-5)=$ ?
(i) $-425$
(ii) 375
(iii) $-625$
(iv) 0
Answer:
(iii) $-625$
Hint: $(50+20)(-20-5)=-(5+20)^{2}=-(25)^{2}=-625$

Question $3 .$
The factors of $x^{2}-6 x+9$ are
(i) $(x-3)(x-3)$
(ii) $(x-3)(x+3)$
(iii) $(x+3)(x+3)$
(iv) $(x-6)(x+9)$
Answer:
(i) $(x-3)(x-3)$
Hint: $x^{2}-6 x+9=x^{2}-2(x)(3)+3^{2}$
$a^{2}-2 a b+b^{2}-(a-b)^{2}=(x-3)^{2}=(x-3)(x-3)$
 

Question $4 .$
The common factors of the algebraic expression $a x^{2} y, b x y^{2}$ and cxyz is
(i) $x^{2} y$
(ii) $x y^{2}$
(iii) $x y z$
(iv) $\mathrm{x}$
Ans:
(iv) $x y$
Hint: $a x^{2} y=a \times x \times x \times y$
$b x y^{2}=b \times x \times y \times y$
$c x y z=C \times x \times y \times z$
Common factor $=\mathrm{xy}$