ML AGGARWAL CLASS 8 CHAPTER 11 Factorisation Exercise 11.3

 Exercise 11.3

Solution 1

Sol :
(i) $\quad x^{2}-12 x+36$

$\Rightarrow(x)^{2}-2 \cdot 6 \cdot x+(6)^{2}$

by using $a^{2}-2 a b+b^{2}=(a-b)^{2}$

$\therefore \quad(x-6)^{2}$
 


(ii) $36 p^{2}-60 p q+25 q^{2}$

=$(6 p)^{2}-2 \cdot 6 p \cdot 5 q+(5 q)^{2}$

=$(6 p-5 q)^{2}$



(iii) $9 x^{2}+66 x y+121 y^{2}$

$(3 x)^{2}+23 x \cdot 11 y+(11 y)^{2}$

$=(3 x+11 y)^{2}$


 
(iv) $a^{4}+6 a^{2} b^{2}+9 b^{4}$

=$\left(a^{2}\right)^{2}+2 \cdot a^{2} \cdot 3 b^{2}+\left(3 b^{2}\right)^{2}$

=$\left(a^{2}+3 b^{2}\right)^{2}$



(v) $x^{2}+\frac{1}{x^{2}}+2$

=$(x)^{2}+2 \cdot x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}$

$=\left(x+\frac{1}{x}\right)^{2}$



Solution 2

Sol :
(i) $4 p^{2}-9$

=$(2 p)^{2}-3^{2}$

by using $a^{2}-b^{2}=(a+b)(a-b)$

=(2 p+3)(2 p-3)


(ii) $4 x^{2}-169 y^{2}$

=$(2 x)^{2}-(13 y)^{2}$

=(2x+13y)(2 x-13 y)


Solution 3

Sol :
(i) $9 x^{2} y^{2}-25$

=$(3 x y)^{2}-5^{2}$

=(3 x y+5)(3 x y-5)


(ii) $16 x^{2}-\frac{1}{144}$

=$(4 x)^{2}-\left(\frac{1}{12}\right)^{2}$

=$\left(4 x^{2}+\frac{1}{12}\right)\left(4 x-\frac{1}{12}\right)$

Solution 4

Sol :
(i) $20 x^{2}-45 y^{2}$

=$5\left(4 x^{2}-9 y^{2}\right)$

=$5\left((2 x)^{2}-(3 y)^{2}\right)$

=5(2 x+3 y)(2 x-3 y)


(ii) $\frac{9}{16}-25 a^{2} b^{2}$

=$\left(\frac{3}{4}\right)^{2}-(5 a b)^{2}$

=$\left(\frac{3}{4}+5 a b\right)\left(\frac{3}{4}-5 a b\right)$

Solution 5

Sol :
(i) $(2 a+3 b)^{2}-16 c^{2}$

=$(2 a+3 b)^{2}-(4 c)^{2}$

=(2 a+3 b-4 c)(2 a+3 b+4 c)


(ii) $1-(b-c)^{2}$

=$1^{2}-(b-c)^{2}$

=(1+b-c)(1-b-c)

Solution 6

Sol :
(i) $9(x+y)^{2}-x^{2}$

=$3^{2}(x+y)^{2}-x^{2}$

=$(3(x+y))^{2}-x^{2}$

=(3(x+y)+x)(3(x+y)-x)

=(3 x+3 y+x)(3 x+3 y-x)

=(4 x+3 y)(2 x+3 y)



(ii) $(2 m+3 n)^{2}-(3 m+2 n)^{2}$

=(2 m+3 n+3 m+2 n)(2 m+3 n-3 m-2 n)

=(5 m+5 n)(n-m)


Solution 7

Sol :
(i) $25(a+b)^{2}-16(a-b)^{2}$

$5^{2}(a+b)^{2}-4^{2}(a-b)^{2}$

$(5 a+5 b)^{2}-(4 a-4 b)^{2}$

=(5a+5 b+4 a-4 b)(5 a+5 b-4 a+4 b)

=(9 a-b)(a+a b)
 


(ii) $9(3 x+2)^{2}-4(2 x-1)^{2}$

=$3^{2}(3 x+2)^{2}-2^{2}(2 x-1)^{2}$

=$(9 x+6)^{2}-(4 x-2)^{2}$

=(9 x+6+4 x-2)(9 x+6-4 x+2)

=(13 x+4)(5 x+7)


Solution 8

Sol :
(i) $x^{3}-25 x$

=$x\left(x^{2}-25\right)$

=$x\left(x^{2}-5^{2}\right)$

=x[(x+5)(x-5)]


(ii) $63 p^{2} q^{2}-7$

=$7\left(9 p^{2} q^{2}-1\right)$

=$7\left((3 p q)^{2}-1^{2}\right)$

=7[(3 p q+1)(3 p q-1)]



Solution 9

Sol :
(i) $32 a^{2} b-72 b^{3}$

=$8 b\left(4 a^{2}-9 b^{2}\right)$

=$8 b\left((2 a)^{2}-(3 b)^{2}\right)$

=8b[(2 a+3 b)(2 a-3 b)]


(ii) $9(a+b)^{3}-25(a+b)$

=$(a+b)\left(3^{2}\left(a+b^{2}\right)^{2}-5^{-2}\right)$

=$(a+b)\left((3 a+3 b)^{2}-5^{2}\right)$

=(a+b)(3 a+3 b+5)(3 a+3 b-5)



Solution 10

Sol :
(i) $x^{2}-y^{2}-2 y-1$

=$x^{2}-\left(y^{2}+2 y+1\right)$           $a^{2}+b^{2}+2 a b=(a+b)^{2}$
                                                                     $(a+b)(a-b)=a^{2}-b^{2}$
=$1 \quad x^{2}-(y+1)^{2}$

=x(x+y+1)(x-y-1)


(ii) $p^{2}-4 p q+4 q^{2}-r^{2}$

=$\left(p^{2}-2 \cdot p \cdot 2 q+(2 q)^{2}\right)-r^{2}$

=$\Rightarrow \quad(p+2 q)^{2}-r^{2}$

=$\Rightarrow \quad(p-2 q+r)(p-2 q-r)$


Solution 11

Sol :
(i) $9 x^{2}-y^{2}+4 y-4$

=$(3 x)^{2}-\left(y^{2}-2 \cdot y \cdot 2+2^{-2}\right)$

=$(3 x)^{2}-(y-2)^{2}$

=(3 x+y-2)(3 x-y+2)



(ii) $4 a^{2}-4 b^{2}+4 a+1$

=$(2 a)^{2}+2 \cdot 2 a \cdot 1+(1)^{2}-(2 b)^{2}$

=$(2 a+1)^{2}-(2 b)^{2}$

=(2 a+1+2 b)(2 a+1-2 b)


Solution 12

Sol :
(i) $625-p^{4}$

=$(25)^{2}-\left(p^{2}\right)^{2}$

=$\left(25+p^{2}\right)\left(25-p^{2}\right)$


(ii) $5 y^{5}-405 y$

=$5 y\left(y^{4}-81\right)$

=$5 y \quad\left(\left(y^{2}\right)^{2}-9^{2}\right)$

=$5 y\left(y^{2}+9\right)\left(y^{2}-9\right)$


Solution 13

Sol :
(i) $x^{4}-y^{4}+x^{2}-y^{2}$

=$\left(x^{4}-y^{4}\right)+\left(x^{2}-y^{2}\right)$

=$\left(\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}\right)+\left(x^{2}-y^{2}\right)$

=$\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)+\left(x^{2}-y^{2}\right)$

=$\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}+1\right)$


(ii) $64 a^{2}-9 b^{2}+42 b c-49 c^{2}$

=$(8 a)^{2}-(3 b)^{2}+7 c(6 b-7 c)$

=(8 a+3 b)(8 a-3 b)+7 c(6 b-7 c)

Comments

Post a Comment

Popular posts from this blog

ML Aggarwal Solution Class 9 Chapter 9 Logarithms MCQs

 MCQs correct Solution from the given four options (1 to 7): Question 1 If log√3 27 = x, then the value of x is (a) 3 (b) 4 (c) 6 (d) 9 Sol : ⇒$\log _{\sqrt{3}} 27=x$ ⇒$(\sqrt{3})^{x}=27$ ⇒$(3)^{\frac{1}{2} x x}=3^{3}$ ⇒$3^{\frac{x}{2}}=3^{3}$ ⇒$\frac{x}{2}=3$ ⇒x=6...(c) Question 2 If log 5 (0.04) = x, then the value of x is (a) 2 (b) 4 (c) -4 (d) -2 Sol : ⇒$\log _{5}(0.04)=x$ ⇒$5^{x}=0.04=\frac{4}{100}=\frac{1}{25}=5^{-2}$ ∴x=-2...(d) Question 3 If log 0.5  64 = x, then the value of x is (a) -4 (b) -6 (c) 4 (d) 6 Sol : ⇒$\log _{0.5} 64=x $ ⇒$0.5^{x}=64$ ⇒$=\left(\frac{1}{2}\right)^{x}=2^{6}$ ⇒$2^{-x}=2^{6}$ ∴-x=6⇒x=-6...(b) Question 4 If $\log _{10} \sqrt[3]{5} x=-3$, then the value of x is (a) $\frac{1}{5}$ (b) $-\frac{1}{5}$ (c) -1 (d) 5 Sol : $\log _{\sqrt[3]{5}} x=-3,(\sqrt[3]{5})^{-3}=x$ $x=\left(5^{\frac{1}{3}}\right)^{-3}=5^{\frac{1}{3}(-3)}=5^{-1}$ $x=\frac{1}{5}$ (b) Question 5 If log (3x + 1) = 2, then the value of x is (a) $\frac{1}{3}$ (b) 99 (c) 33 (d) $\frac{19}{3}$ Sol
Read more

ML Aggarwal Solution Class 9 Chapter 20 Statistics Exercise 20.2

  Exercise 20.2 Question 1 Sol : (i) discrete  (ii) continuous  (iii) discrete  (iv) continuous  (v) continuous  Question 2 Sol : Given data  13, 6, 10 ,5 ,11 ,14 ,2 ,8 ,15 ,16 ,9 ,13 ,17 ,11, 19 ,5 ,7 ,12 ,20 ,21 ,18 ,1 ,8 ,12 ,18 Classes 0-4 5-9 10-14 15-9 20-24 Frequency 2 7 8 6 2 Question 3 Sol : (i)  Classes 1-10 11-20 21-30 31-40 Frequency 8 7 6 6 (ii) Range = Maximum value - Minimum value  =40-2=38 (iii) Third class of frequency table $=\frac{21+30}{2}=\frac{51}{2}$ =25.5 Question 4 Sol : Variate : A particular value of a variable is called variate Class size :  The difference between the actual upper limit and the actual lower li
Read more

ML Aggarwal Solution Class 9 Chapter 3 Expansions Exercise 3.2

Exercise 3.2 Question 1 Sol : Given : x-y=8..(i) xy=5...(ii) Squaring both sides in equation (i) ∴(x-y) 2 =8 2 ⇒x 2 -2.xy+y 2 =64 ⇒x 2 -2(5)+y 2 =64 ⇒x 2 +y 2 =64+10 ⇒x 2 +y 2 =74 Question 2 Sol : ⇒Given : x+y=10...(i) ⇒xy=21....(ii) Squaring both sides in equation (i) ⇒(x+y) 2 =10 2 ⇒x 2 +2xy+y 2 =100 ⇒x 2 +2(21)+y 2 =100 ⇒x 2 +4 2 +y 2 =100 ⇒x 2 +y 2 =100-42 ⇒x 2 +y 2 =58 2(x 2 +y 2 )=2×58=116 Question 3 Sol : ⇒Given : 2a+3b=7..(i) ⇒ab=2....(ii) Squaring both sides in equal (i) ⇒(2a+3b) 2 =7 2 ⇒(2a) 2 +2.2a+3b+(3b) 2 =49 ⇒4a 2 +12ab+9b 2 =49 ⇒4a 2 +12(2)+9b 2 =49 ⇒4a 2 +24+9b 2 =49 ⇒4a 2 +9b 2 =49-24 ⇒4a 2 +9b 2 =25 Question 4 Sol : ⇒Given : 3x-4y=16...(i) ⇒xy=4..(ii) Squaring on both sides in equation (i) ⇒(3x-4y) 2 =16 2 ⇒(3x) 2 -2.3x.4y+(4y) 2 =256 ⇒9x 2 -24xy+16y 2 =256 ⇒9x 2 -24(4)+16y 2 =256 ⇒9x 2 -96+16y 2 =256 ⇒9x 2 +16y 2 =256+96 ⇒9x 2 +16y 2 =352 Question 5 Sol : ⇒Given : x+y=8...(i) ⇒x-y=2...(ii) Since we know that ⇒2(x) 2 +2(y) 2 =(x+y) 2 +(x-y) 2 From (i) ⇒S
Read more