Excercise 10.1
Question 1
(i) Given expression $12 x^{2} y z-4 x y^{2}$
Terms
$12 x^{N} y z$
$-4 x y^{2}$
Numerical cofficient
12
-4
Literal Coefficient
$x^{2} y z$
$x y^{2}$
(ii) Given expression $8+m n+n l-1 m$
Terms
8
mn
nl
-lm
Numerical cofficient
8
1
1
-1
Literal Coefficient
-
mn
nl
lm
(iii) Given expression $\frac{x^{2}}{3}+\frac{4}{6}-x y^{2}$
Terms
$\frac{x^{2}}{3}$
$\frac{y}{6}$
$-x y^{2}$
Numerical co-efficient
$\frac{1}{3}$
$\frac{1}{6}$
$-1$
Literal co - efificient
$x^{2}$
$y$
$x y^{2}$
(iv) Given expression $-4 p+2.3 q+1.7 r$
Terms
$-4 \mathrm{P}$
$2.3q$
$1.7 \mathrm{r}$
Numerical co-efficient
$-4$
$2.3$
$1.7$
Literal co - efificient
$P$
$q$
$r$
Question 2
i) $5 \mathrm{p} \times \mathrm{q} \times \mathrm{r}^{2} \rightarrow$ Monomial
ii) $3 x^{2} \times y \div 2 z \rightarrow$ Monomial
iii) $-3+7 x^{2} \rightarrow$ Binomial
iv) $\frac{5 a^{2}+3 b^{2}+c}{2} \rightarrow$ Trinomial
v) $7 x^{5}-\frac{3 x}{y} \rightarrow$ Binomiol
vi) $5 p \div 3 q-3 p_{x}^{2} \ q_{} \rightarrow$ Binomial
Question 3
(i) $\frac{2}{5} x^{4}-\sqrt{3} x^{2}+5 x-1$
Itis polynomial of degree 4
(ii) $7 x^{3}-\frac{3}{x^{2}}+\sqrt{5}$
due to $-3 x^{-2}$ term, It is not called as polynomiol
$\therefore$ It is Not Pollynomial
(iii) $4 a^{3} b^{2}-3 a b^{4}+5 a b+\frac{2}{3}$
Itis a polynomial of degree 5
(iv) $2 x^{2} y-\frac{3}{x y}+5 y^{3}+\sqrt{3}$
due to negative power in the $-3(x y)^{-1}$
$\therefore$ Itis NoT a Polynomial
Question 4
(i) Arrange terms for column method
$\begin{array}{r}a b-b c \\0+b c-c a \\-a b \quad 0+c a \\\hline 0+0+0\end{array}$
$\therefore a b-b c+b c-c a+c a-a b=0$
(ii) Arrange terms in columns
$5 p^{2} q^{v}+4 p q+7$
$-2 p^{2} q^{2}+9 p q+3$
---------------------------------
$3 p^{v} q^{2}+13 p q+10$
(iii) Arrange terms in columns
$l^{2}+m^{2}+n^{2}+0+0+0$
$0+0+0+\operatorname{lm}+m n+0$
$0+0+0+0+m n+nl$
$0+0+0+1 m+0+n l$
---------------------------------------------------
$l^{2}+m^{2}+n^{2}+2l m+2 m n+2 n l$
(iv) Arrange terms in columns
$4 x^{3}-4 x^{2}+0 x+9$
$^{}+3 x^{2}+5 x+4$
$7 x^{3}+0-11 x+1$
$0+6 x^{2}-13 x+0$
-------------------------------------
$10 x^{3}+2 x^{2}-29 x+14$
Question 5
(i) (diagram should be added)
(ii) $12 x y-3 y z-4 z x+5 x y z$
$8 x y+4 y z+5 z x+0$
$(-) \quad(-) \quad(-)\quad(-)$
--------------------------------------
$4 x y-7 y z-9 z x+5 x y z$
(iii) $5 p_{q}^{2}-2 p q^{2}+5 p q-11 q-3 p+18$
$4 p^{2} q+3 p q^{2}-3 p q+7 q-8 p-10$
$(-) \quad(-)\quad(+) \quad(-) \quad(+)\quad(+)$
-----------------------------------------------------------
$p^{\gamma} q-7 p q^{2}+8 p q-18 q+5 p+28$
Question 6
Horizontal method
$3 x^{2}+5 x y+7 y^{2}+3 \rightarrow 1$
$2 x^{2}-4 x y-3 y^{2}+7 \rightarrow(2)$
$9 x^{2}-8 x y+11 y^{2} \rightarrow(3)$
(3) $-[(1)+(2)]$
$9 x^{2}-8 x y+11 y^{2}-\left[3 x^{2}+5 x y+4 y^{2}+3+2 x^{2}-4 x y-3 y^{2}+7\right]$
$9 x^{2}-8 x y+11 y^{2}-\left[5 x^{2}+x y+4 y^{2}+10\right]$
$9 x^{2}-8 x y+11 y^{2}-5 x^{2}-x y-4 y^{2}-10$
$4 x^{2}-9 x y+7 y^{2}-10$
Question 7
Let
$3 a^{2}-5 a b-2 b^{2}-3---(1)$
$5 a^{2}-7 a b-3 b^{2}+3 a---(2)$
do (2) - (1)
$5 a^{v}-7 a b-3 b^{2}+3 a$
$3 a^{v}-5 a b-2 b^{2}+0-3$
$(-) \quad(+)\quad(+) \quad(-) \quad(+)$
---------------------------------------------------
$2 a^{2}-2 a b-b^{2}+3 a+3$
Question 8
Perimeter of triangle (p) = $7 p^{2}-5 p+11 \rightarrow(1)$
sides
$s_{1}=p^{2}+2 p-1 \rightarrow(2)$
$s_{2}=3 p^{2}-6 p+3 \rightarrow(3)$
$S_{3}=?$
$\begin{aligned} P &=S_{1}+S_{2}+S_{3} \\ S_{3} &=P-\left(S_{1}+S_{2}\right) \end{aligned}$
$-7 p^{2}-5 p+11-\left[p^{2}+2 p-1+3 p^{2}-6 p+3\right]$
$=7 p^{2}-5 p+11-\left[4 p^{2}-4 p+2\right]$
$=7 p^{2}-5 p+11-4 p^{2}+4 p-2$
$S_{3}=3 p^{2}-p+9$
hence third side of triangle
Excercise 10.2
Question 1
(i) $4 x^{3}$ and $-3 x y$
=$4 x^{3} x-3 x y$
=$(4 x-3) \times\left(x^{3} \times x y\right)$
=$-12 x^{4} y$
(ii) $2 x y z$ ond 0
=$(2 x y z) \times 0$
=0
(iii) $-\frac{2}{3} P^{2} q \cdot \frac{3}{4} p q^{2}$ and $5 p q r$
=$\left(\frac{-2}{3} p^{2} q\right) \times\left(\frac{3}{4} p q^{2}\right)(5 p q r)$
=$\frac{-5}{2} p^{4} q^{4} r$
(iv) $-7 a b,-3 a^{3}$ and $\frac{-2}{7} a b^{2}$
$(-7 a b) \times\left(-3 a^{3}\right) \times\left(\frac{-2}{7} a b^{2}\right)$
$\left(-7 x-3 x \times \frac{-2}{7}\right) \times a b \times a^{3} \times a b^{2}$
$-6 a^{5} b^{3}$
(v) $\frac{-1}{2} x^{2},-\frac{3}{5} x y, \frac{2}{3} y z \operatorname{and} \frac{5}{7} x y z$
$\left(-\frac{1}{2} x^{2}\right) \times\left(\frac{-3}{5} x y\right) \times\left(\frac{2}{3} y z\right) \times\left(\frac{5}{7} x y z\right)$
$\left(\frac{-1}{2} \times-\frac{3}{5} \times \frac{2}{3} \times \frac{5}{7}\right) \times x^{2} \times x y \times y z \times x y z$
$\frac{1}{7} x^{4} y^{3} z^{2}$
Question 2
(i) $(3 x-5 y+7 z) \times(-3 x y z)$
=$(3 x x-3 x y z)+(-5 y x-3 x y z)+(7 z x-3 x y z)$
=$-9 x^{2} y z+15 x y^{2} z-21 x y z^{2}$
(ii) $\left(2 p^{2}-3 p q+5 q^{2}+5\right) \times(-2 p q)$
=$\left(2 p^{2} x-2 p q\right)+(-3 p q x-2 p q)+\left(5 q^{2} x-2 p q\right)+$$(5 x-2 p q)$
=$-4 p^{3} q+6 p^{2} q^{2}-10 p q^{3}-10 p q$
(iii) $\left(\frac{2}{3} a^{2} b-\frac{4}{5} a b^{2}+\frac{2}{7} a b+3\right) \times(35 a b)$
$\left(\frac{2}{3} a^{2} b \times 35 a b\right)+\left(\frac{-4}{5} a b^{2} \times 35 a b\right)+\left(\frac{2}{7} a b \times 35 a b\right)+(3 \times 35 a b)$
$\frac{70}{3} a^{3} b-28 a^{2} b^{3}+10 a^{2} b^{2}+105 a b$
(iv) $\left(4 x^{2}-10 x y+7 y^{2}-8 x+4 y+3\right) \times(3 x y)$
$\left(4 x^{2} \times 3 x y\right)+(-10 x y \times 3 x y)+\left(7 y^{2} \times 3 x y\right)+(-8 x \times 3 x y)+(4 y \times 3 x y)+(3 \times 3 x y)$
$12 x^{3} y-30 x^{2} y^{2}+21 x y^{3}-24 x^{2} y+12 x y^{2}+9 x y$
Question 3
(i) Given length (l) = $P^{2} q$
breardth $(b)=p q^{2}$
$\begin{aligned} \text { Rectangle - Area } &=l \times b \\ &=\left(p^{2} q\right) \times\left(p q^{2}\right) \end{aligned}$
Area $=p^{3} q^{3}$
(ii) Given length (l) = 5xy
breadth (b) =$7 x y^{2}$
Rectangle area = L $\times $ b
$\begin{aligned} &=(5 x y) \times\left(7 x y^{2}\right) \\ \text { Area } &=35 x^{2} y^{3} \end{aligned}$
Question 4
(i) Given length (l) = 5ab
breadth(b) = $3 a^{2} b$
height (h) = $7 a^{4} b^{2}$
Volume of rectangular box $=\operatorname{lx} b \times h$
$=5 a b \times 3 a^{2} b \times 7 a^{4} b^{2}$
$=(3 \times 5 \times 7) \times a b \times a^{2} b \times a^{4} b^{2}$
Volume$=105 a^{7} b^{4}$
(ii) Given length (l) = 2pq
breadth (b) = $4 q^{2}$
height $(h)=8 r p$
Volume of retangular box $=\operatorname{lxb} \times h$
$=(2 p q) \times\left(4 q^{2}\right) \times(8 rp)$
$=(2 \times 4 \times 8) P q \times q^{2}+r p$
volume of rectangular box = $64 p^{2} q^{3}r =64 p^{2} q^{3}r $
Question 5
i) $x^{2}\left(3-2 x+x^{2}\right)$
=$3 x^{2}-2 x^{3}+x^{4}$
For x = 1
=$3 \times 1^{2}-2 \times 1^{3}+14$
=$3 \times 1-2 \times 1+1$
=$3-2+1$
=4-2
2 for $x=1$
=$3 x^{2}-2 x^{3}+x^{4}$
=$3 x(-1)^{2}+(-2)(-1)^{3}+(-1)^{4}$
=$3 \times 1+(-2 x-1)+1$
=$3+2+1$
${6}$ for $x=-1$
For $x=\frac{2}{3}$
$3 x^{2}-2 x^{3}+x^{4}$
$3 x\left(\frac{2}{3}\right)^{2}-2\left(\frac{2}{3}\right)^{3}+\left(\frac{2}{3}\right)^{4}$
$3 \times \frac{4}{9}-2 \times \frac{8}{27}+\frac{16}{81}$
$\frac{12}{9}-\frac{16}{27}+\frac{16}{81}$
$\frac{108-48+16}{81}$
$\frac{76}{81}$ for $x=\frac{2}{3}$
$\frac{76}{81}$ for $x=\frac{2}{3}$
For $x=\frac{-1}{2}$
$3 x^{2}-2 x^{3}+x^{4}$
$3 x\left(\frac{-1}{2}\right)^{2}-2 \times\left(-\frac{1}{2}\right)^{3}+\left(-\frac{1}{2}\right)^{4}$
$3 \times \frac{1}{4}-2 \times\left(-\frac{1}{8}\right)+\frac{1}{16}$
$\frac{3}{4}+\frac{2}{8}+\frac{1}{16}$
$\frac{12+4+1}{16}$
$\frac{17}{16}$ for $x=\frac{-1}{2}$
(ii) $5 x y(3 x+4 y-7)-3 y\left(x y-x^{2}+9\right)-8$
$(5 x y \times 3 x)+(5 x y \times 4 y)-5 x y \times 7+(-3 y \times x y)+\left(-3 y x-x^{2}\right)+(-3 y \times 9)-8$
$15 x^{2} y+20 x y^{2}-35 x y-3 x y^{2}+3 x^{2} y-27 y-8$
$18 x^{2} y+17 x y^{2}-62 x y-8$
For $x=2, \quad y=-1$
=$18\left(2^{2}\right)(-1)+\left(17 \times 2 \times(-1)^{2}\right)-(62 \times 2 \times-1)-8$
=$-72+34+124-8$
=78
Question 6
(i) First expression $=4 p\left(2-p^{2}\right)=8 p-4 p^{3}$
Second expression $=8 p^{3}-3 p$
$\quad$ Required $\quad s u m=\left(8 p-4 p^{3}\right) p+\left(8 p^{3}-3 p\right)$
$=4 p^{3}+5 P$
(ii) First expression = $7 x y(8 x+2 y-3)=56 x^{2} y+14 x y^{2}-21 x y $
second expression $=3 y\left(4 x^{2} y-5 x y+8 x y^{2}\right)$
$=12 x^{2} y^{2}-15 x y^{2}+24 x y 3$
Required sum $=\left(56 x^{2} y+14 x y^{2}-21 x y\right)+\left(12 x^{2} y^{2}-15 x y^{2}+\right.$
$\left.24 x y^{3}\right)$
(iii) First expression $=7 x y(8 x+2 y-3)=56 x^{2} y+14 x y^{2}-21 x y$
Second expression $=4 x y^{2}(3 y-7 x+8)=12 x y^{3}-28 x^{2} y^{2}+32 x y^{2}$
Required sum = $\left(56 x^{2} y+14 x y^{2}-21 x y\right)+\left(12 x y^{3}-28 x^{2} y^{2}+\right.$
$\left.32 x y^{+}\right)$
$=12 x y^{3}-28 x^{2} y^{2}+56 x^{2} y+46 x y^{2}-21 x y$
Question 7
(i) First expression = $6 x(x-y+z)-3 y(x+y-z)$
$=6 x^{2}-6 x y+6 x z-3 x y-3 y^{2}+3 y z$
$6 x^{2}+6 x z-6 x y+3 y z-3 y^{2} \rightarrow 1$
Second expression =$2 z(-x+y+z)$
$=-2 x z-2 y z+2 z^{2}-x(2) \rightarrow$
②-①
(-2xz - 2yz + 2$z^{2}$ $-6 x^{2}-6 x z+6 x y-3 y z$ $+3 y 2$)
$=-6 x^{2}+3 y^{2}+2 z^{2}+6 x y-5 y z-8 x z$
(ii) First expression = $7 x y\left(x^{2}-2 x y+3 y^{2}\right)-8 x\left(x^{2} y-4 x y+\right.$ $7 x y^{2}$
$=7 x^{3} y-14 x^{2} y^{2}+21 x y^{3}-8 x^{3} y+32 x^{2} y$ $-56 x^{2} y^{2}$
First expression = $-x^{3} y$ $+21 x y 3$ $-70 x^{2} y^{2}$ $+32 x^{2} y$------①
Second expression = $3 y\left(4 x^{2} y-5 x y+8 x y^{2}\right)$
$=12 x^{2} y^{2}-15 x y^{2}+24 x y^{3}---(2)$
②-①
$=12 x^{2} y^{2}-15 x y^{2}+24 x y^{3}-\left(-x^{3} y+21 x y^{3}-70 x^{2} y^{2}\right.$ $+32 x^{2} y$
=$12 x^{2} y^{2}-15 x y^{2}+24 x y^{3}+x^{3} y-21 x y^{3}+70 x^{2} y^{2}$ $-32 x^{2} y$
=$82 x^{2} y^{2}+x^{3} y+3 x y^{3}-15 x y^{2}-32 x^{2} y$
$=x^{3} y+3 x y^{2}+82 x^{2} y^{2}-15 x y^{2}-32 x^{2} y$
Excercise 10.3
Question 1
(i) $(5 x-2)(3 x+4)$
$5 x(3 x+4)-2(3 x+4)$
$15 x^{2}+20 x-6 x-8$
$15 x^{2}+14 x-8$
(ii) $(a x+b)(c x+d)$
$a x(c x+d)+b(c x+d)$
$a c x^{2}+a d x+b c x+b d$
$a c x^{2}+(a t+b c) x+b d$
(iii) $(4 p-7)(2-3 p)$
$4 P(2-3 p)=7(2-3 p)$
$8 p-12 p^{2}-14+21 P$
$-12 p^{2}+29 p-14$
(iv) $\left(2 x^{2}+3\right)(3 x-5)$
$2 x^{2}(3 x-5)+3(3 x-5)$
$6 x^{3}-10 x^{2}+9 x-15$
$6 x^{3}-10 x^{2}+9 x-15$
(v) $(1.5 a-2.5 b)(1.5 a+2.5 b)$
$1.5 a(1.5 a+2.5 b)-2.5 b(1.5 a+2.5 b)$
$(1.5 \times 1.5) a^{2}+(1.5 \times 2 \cdot 5) a b-(2.5 \times 1.5) a b-(2.5 \times 2.5) b^{2}$
$2.25 a^{2}+3.75 a b-3.75 a b-6.25 b^{2}$
$2.25 a^{2}+0-6.25 b^{2}$
$2 \cdot 25 a^{2}-6.25 b^{2}$
(vi) $\left(\frac{3}{7} p^{2}+4 q^{2}\right)\left(7\left(p^{2}-\frac{3}{4} q^{2}\right)\right.$
$\frac{3}{7} p^{v} \times\left(7 p^{2}-\frac{21}{4} q^{2}\right)+4 q^{2}\left(-1 p^{2}-\frac{21}{4} q^{2}\right)$
$3 p^{4}-\frac{9}{4} q^{2} p^{2}+28 p^{2} q^{2}-21 q^{4}$
$3 p^{4}+\frac{103}{4} p^{2} q^{2}-21 q^{4}$
Question 2
(i) $(x-2 y+3)(x+2 y)$
(daigram should be added)
$x^{2}+3 x+6 y-4 y^{2}$
(ii) $\left(3-5 x+2 x^{2}\right)(4 x-5)$
(daigtam should be added)
Question 3
(i) $\left(3 x^{2}-2 x-1\right)\left(2 x^{2}+x-5\right)$
(diagram should be added)
(ii) $\left(2-3 y-5 y^{2}\right)\left(2 y-1+3 y^{2}\right)$
(diagram should be added)
Question 4
(i) $\left(x^{2}+3\right)(x-3)+9$
$x^{2}(x-3)+3(x-3)+9$
$x^{3}-3 x^{2}+3 x-9+9$
$x^{3}-3 x^{2}+3 x$
(ii) $(x+3)(x-3)(x+4)(x-4)$
$[x(x-3)+3(x-3)][x(x-4)+4(x-4)]$
$\left[x^{2}-3 x+3 x-9\right]\left[x^{2}-4 x+4 x-16\right]$
$\left[x^{2}-9\right]\left[x^{2}-16\right]$
$x^{2}\left(x^{2}-16\right)-9\left(x^{2}-16\right)$
$x^{4}-16 x^{2}-9 x^{2}+144$
$x^{4}-25 x^{4}+144$
(iii) $(x+5)(x+6)(x+7)$
$[(x+5)(x+6)](x+7)$
$[x(x+6)+5(x+6)](x+7)$
$\left(x^{2}+6 x+5 x+30\right)(x+7)$
$\left(x^{2}+6 x+5 x+30\right) x+\left(x^{v}+6 x+5 x+30\right) 7$
$\left(x^{2}+11 x+30\right) x+\left(x^{2}+11 x+30\right) 7$
$x^{3}+11 x^{2}+30 x+7 x^{2}+77 x+210$
$x^{3}+18 x^{2}+107 x+210$
(iv) $(p+q-2 r)(2 p-q+r)-4 q r$
$p(2 p-q+r)+q(2 p-q+r)-2 r(2 p-q+r)-4 q r$
$2 p^{2}-p q+p r+2 p q-q^{2}+q r-4 p y+2 q r-2 r^{2}-4 q r$
$2 p^{2}-q^{2}-2 r^{2}+p q-q r-3 p r$
(v) $(p+q)(r+s)+(p-q)(r-s)-2(p r+q s) .$
$P(r+s)+q(r+s)+p(r-s)-q(r-s)-2 p r-2 q s$
$p r+p s+q r+q s+p r-p s-q r+q s-2 p r-2 q s$
$2 p r-2 p r+p s-p s+q r-q r+2 q s-2 q s$
$0+0+0+0=0$
(vi) $(x+y+z)(x-y+z)+(x+y-z)(-x+y+z)-4 z x$
$x(x-y+z)+y(x-y+z)+z(x-y+z)+x\left(-x+y^{n}+z\right)$
$+y(-x+y+z)-z(-x+y+z)-4 z x$
$x^{2}-x y+x z+x y-y^{v}+y z+x z-y z+z^{2}-x^{2}+x y+x z$
$-x y+y^{2}+y z+x z-y z-z^{2}-4 z x$
$x^{2}-x^{2}+2 x y-2 x y+4 x z-4 x z-y^{N}+y^{2}+2 y z-2 y z$
$+z^{2}-z^{2}$
$0+0+0+0+0+0=0$
Question 5
Sides of rectangles $s_{1}=5 x^{2}+25 x y+4 y^{2}$
$S_{2}=2 x^{2}-2 x y+3 y r$
Area of rectangle $=S_{1} \times S_{2}$
$A=\left(5 x^{2}+25 x y+4 y^{2}\right)\left(2 x^{2}-2 x y+3 y^{2}\right)$
$A=10 x^{4}+40 x^{3} y-27 x^{2} y^{2}+67 x y^{3}+12 y 4$
(diagram should be added)
Excercise 10.4
Question 1
(i) $-39 p q^{2} r^{5} \div-24 p^{3} q^{3} r$
$=\frac{-39 p q^{x}+5}{-24 p^{3} q^{3} r}$
$\frac{13 \cdot r^{4}}{8 p^{2} q}$
(ii) $-\frac{3}{4} a^{r} b^{3} \div \frac{6}{7} a^{3} b^{2}$
=$\frac{\frac{-3}{4} a^{2} b^{3}}{\frac{6}{7} a^{3} b^{2}}$
$=\frac{-3}{4} \times \frac{7}{6}-\frac{b}{a}$
$=\frac{-7 b}{8 a}$
Question 2
(i)
$3 x^{3}-\frac{8}{3} x^{2}-4$
$\begin{array}{l}3 x \sqrt{9 x^{4}-8 x^{3}-12 x+3} \\\frac{\left(-9 x^{4}\right.}{0-8 x^{3}} \\\frac{(+)^{-8 x^{3}}}{0-12 x} \\\frac{\frac{1-12 x}{}}{0+3}\end{array}$
Quotient $=3 x^{3}-\frac{8}{3} x^{2}-4 ;$ Remainder $=3$
(ii) (daigram should be added)
Quotient =3 x+5, Remainder =0
Question 3
(i) (daigram should be added)
Quotient =3 x+5 , Remainder =0
(ii) (daigram should be added)
Quotient $=y^{2}$-y-1 Remainder =2
(iii) (daigram should be added)
Qustient =-2 x+3, Remainder =2
(iv) (daigram should be added)
Quotient = $x^{2}-4 x+4$, Remainder =0
Question 4
(i) (daigram should be added)
Quotient $=2 x^{2}+5 x+3$, Remainder $=-4$
(ii) (daigram should be added)
Quotient $=m^{2}-5 m-5$, Remainder =2
Question 5
(i) (daigram should be added)
Quotient =a+1, Remainder =0
(ii) (daigram should be added)
Quotient =4 x-3, Remainder =-3
Question 6
Given area of rectangle = $8 x^{2}-45 y^{2}+18 x y$
one sitle $S_{1}=4 x+15 y$
other side $\quad S_{2}=?$
$A=S_{1} \times S_{2}$
$S_{2}=A \div S_{1}$
(daigram should be added)
Quatient =2 x-3 y, Remainder =0
∴ length other side of rectangle
$s_{2}=2 x-3 y$
Excercise 10.5
Question 1
(i) $(3 x+5)(3 x+5)$
$\begin{aligned}(3 x+5)^{2} &=(3 x)^{2}+2(3 x) .5+5^{2}\left(\therefore(a+b)^{2}=a^{2}+2 a b+b^{2}\right) \\ &=9 x^{2}+30 x+25 \end{aligned}$
(ii) $\begin{aligned}(9 y-5) &(9 y-5) \\(9 y-5)^{2} &=(9 y)^{2}-2(9 y) 5+(+5)^{2}\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right) \\ &=81 y^{2}-90 y+25 \end{aligned}$
(iii) $(4 x+11 y)(4 x-11 y)$
=$(4 x)^{2}-(11 y)^{2}\left(\because(a+b)(a-b)=a^{2}-b^{2}\right)$
=$16 x^{2}-121 y^{2}$
(v) $\left(\frac{2}{a}+\frac{5}{b}\right)\left(\frac{2}{a}+\frac{5}{b}\right)$
$\begin{aligned}\left(\frac{2}{a}+\frac{5}{b}\right)^{2} &=\left(\frac{2}{a}\right)^{2}+2 \cdot \frac{2}{a} \cdot \frac{5}{b}+\left(\frac{5}{b}\right)^{2}\left(\because \cdot(a+b)^{2}=a^{2}+2 a b+b^{2}\right) \\ &=\frac{4}{a^{2}}+\frac{20}{a b}+\frac{25}{b^{2}} \end{aligned}$
(vi) $\left(\frac{p^{2}}{2}+\frac{2}{q^{2}}\right)\left(\frac{p^{2}}{2}-\frac{2}{q^{2}}\right)$
$\left(\frac{p^{2}}{2}\right)^{2}-\left(\frac{2}{q^{2}}\right)^{2} \quad\left(\because(a+b)(a-b)=a^{2}-b^{2}\right.$
$\frac{p^{4}}{4}-\frac{4}{q^{4}}$
Question 2
(i)
$\begin{aligned} 81^{2} &=(80+1)^{2} \\ &=80^{2}+2.80 .1+1^{2}\left(\because(a+b)^{2}=a^{2}+2 a b+b^{2}\right) \\ &=6400+160+1 \\ 81^{2} &=6561 \end{aligned}$
(ii)
$\begin{aligned} 97^{2} &=(100-3)^{2} \\ &=(100)^{2}-2 \cdot 100 \cdot 3+3^{2}\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right] \\ &=10000-600+9 \\ &=9409^{} \end{aligned}$
(iii)
$\begin{aligned} 105^{2} &=(100+5)^{2} \\ &=100^{2}+2.100 \cdot 5+5^{2}\left(\cdots(a+b)^{2}=a^{2}+2 a b+b^{2}\right) \\ &=10000+1000+25 \\ &=11025 \end{aligned}$
(iv)
$\begin{aligned} 997^{2} &=(1000-3)^{2} \\ &=(1000)^{2}-2 \cdot 1000 \cdot 3+3^{2}\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right) \\ &=1000000-6000+9 \\ 997^{2} &=994009 \end{aligned}$
(v)
$\begin{aligned} 6.1^{2} &=(6+0.1)^{2} \\ &=6^{2}+2 \cdot 6 \cdot(0.1)+0.1^{2}\left(\because(a+b)^{\gamma}=a^{2}+2 a b+b^{2}\right) \\ &=36+1.2+0.01 \\ 6.1^{2} &=31.21 \end{aligned}$
(vi)
$\begin{aligned} 496 \times 504 &=(500-4)(500+4) \\ &=(500)^{2}-4^{2}\left(\because \cdot(a+b)(a-b)=a^{2}-b^{2}\right) \\ &=250000-16 \\ 496 \times 504 &=249984 \end{aligned}$
(vii)
$\begin{aligned} 20.5 \times 19.5 &=(20+0.5)(20-0.5) \\ & \left.=20^{2}-0.5^{2}(19+b)(a-b)=a^{2}-b^{2}\right) \\ &=400-0.25 \\ 20.5 \times 19.5 &=399.75 \end{aligned}$
(viii)
$\begin{aligned} 9.6^{2} &=(10-0.4)^{2} \\ &=10^{2}-2.10 \cdot(0.4)+(0.4)^{2}\left(\because \cdot(a-b)^{2}=a^{2}-2 a b+b^{2}\right) \\ &=100-8+0.16 \\ 9.6^{2} &=92.16 \end{aligned}$
Question 3
(i)
$\begin{aligned}(p q+5 r)^{2} &=(P q)^{2}+2 \cdot p q \cdot 5 r+(5 r)^{2}\left(-(a+b)^{2}=a^{2}+b^{2}+2 a b\right) \\ &=p^{2} q^{r}+10 p q r+25 2^{2} \end{aligned}$
(ii)
$\begin{aligned}\left(\frac{5}{2} a-\frac{3}{5} \cdot b\right)^{2} &=\left(\frac{5}{2} \cdot a\right)^{2}-2 \cdot \frac{5}{2} \cdot a \cdot \frac{3}{5} \cdot b+\left(\frac{3}{5} b\right)^{2} \\ &\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right) \\ &=\frac{25}{4} a^{2}-3 a b+\frac{9}{25} b^{2} \end{aligned}$
(iii)
$\begin{aligned}(\sqrt{2} \cdot a+\sqrt{3} \cdot b)^{2} &=(\sqrt{2} \cdot a)^{2}+2 \sqrt{2} \cdot a \cdot \sqrt{3} \cdot b+(\sqrt{3} \cdot b)^{2} \\(\therefore& \left.(a+b)^{2}=a^{2}+2 a b+b^{2}\right) \\ &=2 a^{2}+2 \sqrt{6} a b+3 b^{2} \end{aligned}$
(iv)
$\left(\frac{2 x}{3 y}-\frac{3 y}{2 x}\right)^{2}=\left(\frac{2 x}{3 y}\right)^{2}-2 \cdot \frac{2 x}{3 y} \cdot \frac{3 y}{2 x}+\left(\frac{3 y}{2 x}\right)^{2}$ $\left(∴(a+b)^{2}=a^{2}+b^{2}+2 a b\right)$
$=\frac{4 x^{2}}{9 y^{2}}-2+\frac{9 y^{2}}{4 x^{2}}$
Question 4
(i)
$\begin{aligned}(x+7)(x+3) &=x^{2}+(7+3) x+7 \times 3\left(:(x+a)(x+b)=x^{2}+(a+b) x+a b\right) \\ &=x^{2}+10 x+21 \end{aligned}$
(ii) $(3 x+4)(3 x-5)=(3 x)^{v}+(4+(-5))(3 x)+4 x-5$
$\left(\because(x+a)(x+b)=x^{2}+(a+b) x+a b\right)$
$=9 x^{2}-3 x-20$
(iii) $\left(p^{2}+2 q\right)\left(p^{2}-3 q\right)=\left(p^{2}\right)^{2}+(2 q+(-3 q)) p^{2}+2 q x-3 q$
$\left(\because(x+a)(x+b)=x^{2}+(a+b) x+a b\right)$
$=p^{4}-^{2} q^{2}-6 q^{2}$
$=p^{4}-p_{q}^{2}-6 q^{2}$
(iv) $(a b c+3)(a b c-5)=(a b c)^{2}+(3+(-5)) \cdot a b c+3 x-5$
$\left(\because(x+a)(x+b)=x^{2}+(a+b) \cdot x+a b\right)$
$=(a b c)^{2}-2 a b c-15$
Question 5
(i) $203 \times 204=(200+3)(200+4)$
$=(200)^{2}+(3+4) 200+3 \times 4\left(\because(x+a)(x+b)=x^{2}+(a+b)+x+ab\right.$
$=40000+1400+12$
$=41412$
(ii) $8.2 \times 8.7=(8+0.2)(8+0.7)$
$=8^{2}+(0.2+0.7) 8+0.2 \times 0.7$
$=64+7.2+0.14$
$=71.34$
(iii) $107 \times 93=(100+7)(100-7)$
$=(100)^{2}+(7+(-7)) \cdot 100+7 x-7$
$\left(\because(x+a)(x+b)=x^{2}+(a+b) \cdot x+a b\right)$
$=10000+0.100=49$
$=9951$
Question 6
(i) $53^{2}-47^{2}=(53+47)(53-47)$ $\left(∴ a^{2}-b^{2}=(a+b)(a-b)\right)$
$=(100)(6)$
$=600$
(ii) $(2.05)^{2}-(0.95)^{2}=(2.05+0.95)(2.05-0.95)$
$=3 \times 0.1$
$=0.3$
Question 7
i. $(2 x+5 y)^{2}+(2 x-5 y)^{2}$
$(2 x)^{2}+(5 y)^{2}+2 \cdot 2 x \cdot 5 y+(2 x)^{2}+(5 y)^{2}-2 \cdot 2 x \cdot 5 y$
$\because(a+b)^{2}=a^{2}+b^{2}+2 a b$
$2(2 x)^{2}+2(5 y)^{2}$
$2 \cdot\left[4 x^{-1}+25 y^{2}\right]$
$8 x^{2}+50 y^{2}$
(ii) $\left(\frac{7}{2} a-\frac{5}{2} b\right)^{2}-\left(\frac{5}{2} a-\frac{z}{2} b\right)^{2}$
=$\left(\frac{7}{2} a\right)^{2}+\left(\frac{5}{2} b\right)^{2}-2 \cdot \frac{7}{2} a \cdot \frac{5}{2} \cdot b-\left[\left(\frac{5}{2} a\right)^{n}+\left(\frac{7}{2} b\right)^{2}-2 \cdot \frac{5}{2} \cdot a \cdot \frac{7}{2} b\right]$
=$\frac{49}{4} a^{2}+\frac{25}{4} b^{2}-2 \cdot \frac{7}{2} \cdot \frac{5}{2} \cdot b-\frac{25}{4} a^{2}-\frac{49}{4} b^{2}+2 \cdot \frac{5}{2} a\cdot \frac{7}{2} \cdot b$
=$\left(\frac{49}{4}-\frac{25}{4}\right) a^{2}+\left(\frac{25}{4}-\frac{49}{4}\right) b^{2}$
=$\frac{24}{4} \cdot a^{2}-\frac{24}{4} \cdot b^{2}$
=$6\left(a^{2}-b^{2}\right)$
(iii) $\left(p^{2}-qr^{2} \cdot \right)^{2}+2 p^{2} q^{2} 2$
$\left(p^{2}\right)^{2}-2 \cdot p^{2} \cdot q^{2} \cdot 2+\left(q^{2} \cdot 2\right)^{2}+2 p^{2} q^{2} \cdot r$ $\left(\therefore(a+b)^{2}=a^{2}+2 a b+b^{2}\right.$
$p^{4}-2 p^{2} q^{2} \cdot r+q^{4} \cdot r^{2}+2 p^{2} q^{2} \cdot r$
$p^{4}+q^{4} \cdot r^{2}$
Question 8
L H S
(i) $(4 x+7 y)^{2}-(4 x-7 y)^{n}$
$(4 x)^{2}+(7 y)^{2}+2 \cdot 4 x \cdot 7 y-\left[(4 x)^{2}+(7 y)^{2}-2.4 x-7 y\right]$
$\left(\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right)$
$(4 x)^{2}+(7 y)^{2}+2.4 x \cdot 7 y-(4 x)^{2}-(7 y)^{2}+2 \cdot 4 x \cdot 7 y$
$4.4 x \cdot 7 y$
$112 x y$= R.H.S
(ii) $\left(\frac{3}{7} p-\frac{7}{6} q\right)^{2}+p q$
$\left(\frac{3}{7} p\right)^{2}+\left(\frac{7}{6} q\right)^{2}-2 \cdot \frac{3}{7} p \cdot \frac{1}{6} \cdot q+p q$
$\left(\because(a-b)^{2}=a^{2}+b^{2}-2 a b\right)$
$\frac{9}{49} p^{2}+\frac{49}{36} q^{2}-p q+p q$
$\begin{aligned} \frac{9}{49} \cdot p^{2}+\frac{49}{36} q^{2} &=R \cdot H \cdot S \\ \therefore \quad L . H \cdot S &=R \cdot H \cdot S \end{aligned}$
(iii) $L \cdot H \cdot s=(p-q)(p+q)+(p-r)(q+r)+(r-p)(r+p)$
$=p^{2}-q^{2}+q^{2}-r^{2}+r^{2}-p^{2}$
$\left(\because+(a+b)(a-b)=a^{2}-b^{2}\right]$
$=0=R . H \cdot S$
$\therefore \quad L H S=R H S$
Question 9
given $\left(x+\frac{1}{x}\right)=2$
Squaring on both sides
(i) $\left(x+\frac{1}{x}\right)^{2}=2^{2}$
$x^{2}+2 \cdot x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=$$4\left(\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right)$
$x^{2}+2+\frac{1}{x^{2}}=4$
$x^{2}+\frac{1}{x^{2}}=4-2$
$x^{2}+\frac{1}{x^{2}}=2$
(ii) Again squaring on both sides
$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=2^{2}$
$\left(x^{2}\right)^{n}+2 \cdot x^{2}-\frac{1}{x^{2}}+\left(\frac{1}{x^{2}}\right)^{2}=4$
$x^{4}+2+\frac{1}{x^{4}}=4$
$x^{4}+\frac{1}{x^{4}}=4-2$
$x^{4}+\frac{1}{x^{4}}=2$n
Question 10
(i) $x-\frac{1}{x}=7$
Squaring on boths sides
$\left(x-\frac{1}{x}\right)^{2}=7^{2}$
$x^{2}-2 \cdot x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}=49$
$\left(\because(a-b)^{2}=a^{2}-2 a b+b^{2}\right)$
$x^{2}-2+\frac{1}{x^{2}}=49$
$x^{2}+\frac{1}{x^{2}}=49+2$
$x^{2}+\frac{1}{x^{2}}=51$
(ii) $x^{2}+\frac{1}{x^{2}}=51$
squaring on both sides
$\left(x^{2}+\frac{1}{x^{2}}\right)^{2}=51^{2}$
$\left(x^{2}\right)^{2}+2 \cdot x^{2} \cdot \frac{1}{x^{2}}+\left(\frac{1}{x^{2}}\right)^{2}=2601$
$x^{4}+2+\frac{1}{x^{4}}=2601$
$x^{4}+\frac{1}{x^{4}}=2601-2$
$x^{4}+\frac{1}{x^{4}}=2599$
Question 11
$x^{2}+\frac{1}{x^{2}}=23$
(i) $x^{2}+\frac{1}{x^{2}}=23$
Adding 2 ' on bolb sides
$x^{2}+\frac{1}{x^{2}}+2=23+2$
$(x)^{2}+\left(\frac{1}{x}\right)^{2}+2 \cdot x \cdot \frac{1}{x}=25\left(\because a^{2}+b^{2}+2 a b=(a+b)^{2}\right)$
$\left(x+\frac{1}{x}\right)^{2}=25$
$x+\frac{1}{x}=5$
(ii) $x+\frac{1}{x^{2}}=25^{2}$
Substract '2' on both sides
$x^{2}+\frac{1}{x^{2}}-2=23-2$
$(x)^{2}+\left(\frac{1}{x}\right)^{n}-2 \cdot x \cdot \frac{1}{x}=21$
$\left(x-\frac{1}{x}\right)^{2}=21\left(\because \quad a^{2}+b^{2}-2 a b=(a-b)^{2}\right)$
$x-\frac{1}{x}=\sqrt{21}$
$x-\frac{1}{x}=3 \sqrt{3}$
Question 12
given $\quad a+b=9, \quad a b=10$
Squaring on both sides
$(a+b)^{2}=9^{2}$
$a^{2}+b^{2}+2 a b=81$
$a^{2}+b^{2}+2 \times 10=81 \quad(\because$ given $a b=10)$
$a^{2}+b^{2}+20=81$
$a^{2}+b^{2}=61$
Question 13
given $a-b=6, a^{2}+b^{2}=42$
$a-b=6$
Squaring on both Sides
$(a-b)^{2}=6^{2}$
$a^{2}+b^{2}-2 a b=36$
$42-2 a b=36\left(\because a^{2}+b^{2}=42\right)$
$42-36=2 a b$
$2 a b=6$
$a b=3$
Question 14
given $a^{2}+b^{2}=41, a b=4$
(i) Consider
$\begin{aligned}(a+b)^{2} &=a^{2}+b^{2}+2 a b \\(a+b)^{2} &=41+2 \times u=41+8 \\(a+b)^{2} &=49 \\ a+b &=7 \end{aligned}$
(ii) Consider
$\begin{aligned}(a-b)^{2} &=a^{2}+b^{2}-2 a b \\ &=41-2 \times 4=41-8=33 . \\(a-b)^{2} &=33 \\ a-b &=\sqrt{33} \end{aligned}$
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