KC Sinha: Exercise 4.4 - Mathematics Solution Class 9 Chapter 4 बीजीय सर्वसमिकाएँ
KC Sinha: Exercise 4.4 - Mathematics Solution Class 9 Chapter 4 बीजीय सर्वसमिकाएँ

Question 1

उपयुक्त सर्वसमिकाओ का उपयोग कर निम्नलिखित गुणनफल ज्ञात कीजिए
(i) (x+3)(x+3)
Sol :
Here we use identity
(x+a)(x+a)=x2+(a+b)x+ab
a=3 , b=3
⇒x2+(3+3)x+3×3
⇒x2+6x+9

(ii) (x-3)(x+5)
Sol :
Using identity:
(x+a)(x+b)=x2+(a+b)x+ab
a=-3 , b=5
⇒x2+(-3+5)x+(-3)(5)
⇒x2+2x-15

(iii) (5x+2)(5x-3)
Sol:
Using identity:
(x+a)(x+b)=x2+(a+b)x+ab
replace x=5x , a=2 , b=-3
⇒(5x)2+(2-3)5x+(2)(-3)
⇒25x2-5x-6

(iv) $\left(y^2-\dfrac{5}{2}\right)\left(y^2+\dfrac{7}{2}\right)$
Sol :
Using identity :
(x+a)(x+b)=x2+(a+b)x+ab
Replace x=y2 , $a=-\dfrac{5}{2},b=\dfrac{7}{2}$
⇒$(y^2)^2+\left(-\dfrac{5}{2}+\dfrac{7}{2}\right)y^2+\left(-\dfrac{5}{2}\right)\left(\dfrac{7}{2}\right)$
⇒$y^4+\left(\dfrac{-5+7}{2}\right)y^2-\dfrac{35}{4}$
⇒$y^4+y^2-\dfrac{35}{4}$

(v) (2x-1)(2x+5)
Sol :
Using identity :
(x+a)(x+b)=x2+(a+b)x+ab
Replace x=2x , a=-1 ,b=5
⇒(2x)2+(-1+5)2x+(-1)(5)
⇒4x2+8x-5

(vi) (5+2x)(5-2x)
Sol :
Using identity :
a2-b2=(a+b)(a-b)
Replace a=5 ,b=2x
⇒52-(2x)2
⇒25-4x2

(vii) $\left(y-\dfrac{x}{10}\right)\left(y+\dfrac{x}{10}\right)$
Sol:
Using identity :
a2-b2=(a+b)(a-b)
Replace $a=y ,b=\dfrac{x}{10}$
⇒$y^2-\left(\dfrac{x}{10}\right)^2$
⇒$y^2-\dfrac{x^2}{100}$

(viii) $\left(x+\dfrac{3}{5}\right)\left(x-\dfrac{3}{5}\right)$
Sol :
Using identity :
a2-b2=(a+b)(a-b)
Replace $a=x ,b=\dfrac{3}{5}$
⇒$x^2-\left(\dfrac{3}{5}\right)^2$
⇒$x^2-\dfrac{9}{25}$

Question 2

सीधे गुणा किए बिना निम्नलिखित गुणनफलो के मान ज्ञात कीजिए:
(i) 105×106
Sol :
$\dfrac{105+106}{2}=105.5$

(i) 105×106
Sol :
$\dfrac{105+106}{2}=105.5$
Hence , we can express 105 and 106 in terms of 105.5 as follows
⇒105×106
⇒(105.5-0.5)(105.5+0.5)
Using identity :
a2-b2=(a+b)(a-b)
बदलन के a=105.5 ,b=0.5
⇒(105.5)2-(0.5)2
⇒11130.25-0.25
⇒11130

ALTERNATE METHOD
⇒105×106=(100+5)(100+6)

=(100)2+(5+6)100+5×6

=10000+1100+30
=11130

(ii) 103×98
Sol :
$\dfrac{103+98}{2}=100.5$
Hence , we can express 103 and 98 in terms of 100.5 as follows
⇒103×98
⇒(100.5+2.5)(100.5-2.5)
Using identity :
a2-b2=(a+b)(a-b)
बदलन के a=100.5 ,b=2.5
⇒(100.5)2-(2.5)2
⇒10100.25-6.25
⇒10094

(iii) 97×98
Sol :
$\dfrac{97+98}{2}=97.5$
Hence , we can express 97 and 98 in terms of 97.5 as follows
⇒97×98
⇒(97.5-0.5)(97.5+0.5)
Using identity :
a2-b2=(a+b)(a-b)
बदलन के a=97.5 ,b=0.5
⇒(97.5)2-(0.5)2
⇒9506.25-0.25
⇒9506

Question 3

निम्नलिखित को प्रसारित रुप मे लिखिए।
(i) (3a+4b+5c)2
Sol :
Using identity :
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
Replace a=3a,b=4b ,c=5c
⇒(3a)2+(4b)2+(5c)2+2(3a)(4b)+2(4b)(5c)+2(5c)(3a)
⇒9a2+16b2+25c2+24ab+40bc+30ac

(ii) (4a-2b-3c)2
Sol :
Using identity :
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
Replace a=4a,b=-2b,c=-3c
⇒(4a)2+(-2b)2+(-3c)2+2(4a)(-2b)+2(-2b)(-3c)+2(-3c)(4a)
⇒16a2+4b2+9c2-16ab+12bc-24ac

(iii) (2x-y+z)2
Sol :
Using identity :
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
Replace a=2x,b=-y,c=-z
⇒(2x)2+(-y)2+(-z)2+2(2x)(-y)+2(-y)(-z)+2(-z)(2x)
⇒4x2+y2+z2-4xy-2yz+4zx

(iv) (2x+3y+z)2
Sol :
Using identity :
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
Replace a=2x,b=3y,c=z
⇒(2x)2+(3y)2+(z)2+2(2x)(3y)+2(3y)(z)+2(z)(2x)
⇒4x2+9y2+z2+12xy+6yz+4zx

(v) $\left(\dfrac{a}{3}-\dfrac{b}{2}+2\right)^2$
Sol :
Using identity :
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
Replace $a=\dfrac{a}{3},b=-\dfrac{b}{2},c=2$
⇒$\left(\dfrac{a}{3}\right)^2+\left(-\dfrac{b}{2}\right)^2+(2)^2+2\left(\dfrac{a}{3}\right)\left(-\dfrac{b}{2}\right)+2\left(-\dfrac{b}{2}\right)(2)+2(2)\left(\dfrac{a}{3}\right)$
⇒$\dfrac{a^2}{9}+\dfrac{b^2}{4}+4-\dfrac{ab}{3}+\dfrac{4}{3}a-2b$

(vi) (-2x+5y-3z)2
Sol :
Using identity :
(a+b+c)2=a2+b2+c2+2ab+2bc+2ca
Replace a=-2x, b=5y, c=-3z
⇒(-2x)2+(5y)2+(-3z)2+2(-2x)(5y)+2(5y)(-3z)+2(-3z)(-2x)
⇒4x2+25y2+9z2-20xy-30yz+12zx

Question 4

निम्नलिखित घनो को प्रसारित रुप मे लिखिए
(i) (3a+4b)3
Sol :
Using identity:
(x+y)3=x3+y3+3x2y+3xy2
Replace x=3a , y=4b
⇒(3a)3+(4b)3+3(3a)2(4b)+3(3a)(4b)2
⇒27a3+64b3+108a2b+144ab2

(ii) (5p-3q)3
Sol :
Using identity:
(x-y)3=x3-y3-3x2y+3xy2
Replace x=5p , y=3q
⇒(5p)3+(3q)3+3(5p)2(3q)+3(5p)(3q)2
⇒125p3-225p2q+135pq2-27q3

(iii) $\left(3p-\dfrac{1}{6}\right)^3$
Sol :
Using identity:
(x-y)3=x3-y3-3x2y+3xy2
Replace x=3p , $y=\dfrac{1}{6}$
⇒$(3p)^3-\left(\dfrac{1}{6}\right)^3-3(3p)^2\left(\dfrac{1}{6}\right)+3(3p)\left(\dfrac{1}{6}\right)^2$
⇒$27p^3-\dfrac{1}{216}-\dfrac{9}{2}p^2+\dfrac{1}{4}p$

(iv) (4a-3b)3
Sol :
Using identity:
(x-y)3=x3-y3-3x2y+3xy2
Replace x=3p , $y=\dfrac{1}{6}$
⇒$(4a)^3-\left(\dfrac{1}{6}\right)-3(3p)^2\left(\dfrac{1}{6}\right)+3(3p)\left(\dfrac{1}{6}\right)^2$
⇒64a3-27b3-144a2b+108ab2

Question 5

उपयुक्त सर्वसमिकाओ का प्रयोग कर, निमनलिखित मे से प्रत्येक का मान ज्ञात कीजिए
(i) (104)3
Sol :
=(100+4)3
Using identity:
(x+y)3=x3+y3+3x2y+3xy2
Replace x=100 , y=4
⇒(100)3+(4)3+3(100)2(4)+3(100)(4)2
⇒1000000+64+3×10000×4+3×100×16
⇒1124864

(ii) (999)3
Sol :
=(1000-1)3
Using identity:
(x-y)3=x3-y3-3x2y+3xy2
Replace x=1000 , y=1
⇒(1000)3-(1)3-3(1000)2(1)+3(1000)(1)2
⇒1000000000-1-3000000+3000
⇒997002999