RD Sharma Solutions Class 9 Algebraic Identities Exercise 4.1

RD Sharma Solutions Class 9 Chapter 4 Ex 4.1

1) Evaluate each of the following using identities:

(i) $(2x-\frac{1}{x})^{2}$

Solution:

It is given,

$(2x-\frac{1}{x})^{2}=(2x)^{2}+(\frac{1}{x})^{2}-2*2x*\frac{1}{x}$

$(2x-\frac{1}{x})^{2}=4x^{2}+\frac{1}{x^{2}}-4\;\;\;\;\;\;[because (a-b)^{2}=a^{2}+b^{2}-2ab\;]$

here, $a=2x,\;b=\frac{1}{x}$

$Therefore, (2x-\frac{1}{x})^{2}=4x^{2}+\frac{1}{x^{2}}-4$

(ii) (2x+y) (2x-y)

Solution:

It is given, (2x+y) (2x-y)

= $(2x)^{2}-(y)^{2}\;\;\;\;\;\;\;[because (a+b)(a-b)=a^{2}-b^{2}]$

= $4x^{2}-y^{2}$

$Therefore, (2x+y) (2x-y)=4x^{2}-y^{2}$

(iii) $(a^{2}b-ab^{2})^{2}$

Solution:

It is given,$(a^{2}b-ab^{2})^{2}$

$=(a^{2}b)^{2}+(ab^{2})^{2}-2*a^{2}b*ab^{2}\;\;\;\;\;\;\;[because (a-b)^{2}=a^{2}+b^{2}-2ab]$

here, $a=a^{2}b,\;b=ab^{2}$

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

$Therefore,(a^{2}b-ab^{2})^{2}=a^{4}b^{2}+b^{4}a^{2}-2a^{3}b^{3}$

(iv) (a-0.1) (a+0.1)

Solution:

It is given,(a-0.1) (a+0.1)

$=a^{2}-(0.1)^{2}\;\;\;\;\;\;\;[because (a+b)(a-b)=a^{2}-b^{2}]$

here a = a and b = 0.1

$=a^{2}-0.01$

$Therefore, (a-0.1)(a+0.1)=a^{2}-0.01$

(v) $(1.5x^{2}-0.3y^{2})(1.5x^{2}+0.3y^{2})$

Solution:

It is given,$(1.5x^{2}-0.3y^{2})(1.5x^{2}+0.3y^{2})$

$=(1.5x^{2})^{2}-(0.3y^{2})^{2}\;\;\;\;\;\;\;[because (a+b)(a-b)=a^{2}-b^{2}]$

here, $a=1.5x^{2},\;b=0.3y^{2}$

$=2.25x^{4}-0.09y^{4}$

$Therefore, (1.5x^{2}-0.3y^{2})(1.5x^{2}+0.3y^{2})=2.25x^{4}-0.09y^{4}$

2) Evaluate each of the following using identities:

(i) $(399)^{2}$

Solution:

As we have,

3992 = (400-1)2

= (400)2+(1)2 – 2x400x1               [ because(a-b)2 = a2+ b2-2ab ]

here, a = 400 and b = 1

= 160000 + 1 – 8000

= 159201

So, (399)2 = 159201.

(ii) (0.98)2

Solution:

As we have,

(0.98)2 = (1-0.02)2

= (1)2 + (0.02)2 – 2x1x0.02

= 1 + 0.0004 – 0.04                 [ Where, a=1 and b=0.02 ]

= 1.0004 – 0.04

= 0.9604

So, (0.98)2 = 0.9604

(iii) 991×1009

Solution:

As we have,

991×1009

= (1000-9) (1000+9)

= (1000)2 – (9)2                       [because (a+b) (a-b) = a2 – b2 ]

= 1000000 – 81                       [ Where a=1000 and b=9 ]

= 999919

So, 991×1009 = 999919

(iv) 117×83

Solution:

As we have,

117×83

= (100+17) (100-17)

= (100)2 – (17)2                       [ because (a+b) (a-b) = a2 – b2 ]

= 10000 – 289             [ Where a=100 and b=17 ]

= 9711

So, 117×83 = 9711

3) Simplify each of the following:

(i) 175 x 175 +2 x 175 x 25 + 25 x 25

Solution:

As we have,

175 x 175 +2 x 175 x 25 + 25 x 25 = (175)2 + 2 (175) (25) + (25)2

= (175+25)2                 [because  a2+ b2+2ab = (a+b)2 ]

= (200)2                       [ Where a=175 and b=25 ]

= 40000

So, 175 x 175 +2 x 175 x 25 + 25 x 25 = 40000.

(ii) 322 x 322 – 2 x 322 x 22 + 22 x 22

Solution:

As we have,

322 x 322 – 2 x 322 x 22 + 22 x 22

= (322-22)2                  [because  a2+ b2-2ab = (a-b)2 ]

= (300)2                       [ Where a=322 and b=22 ]

= 90000

So, 322 x 322 – 2 x 322 x 22 + 22 x 22= 90000.

(iii) 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24

Solution:

As we have,

0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24

= (0.76+0.24)2                         [ because a2+ b2+2ab = (a+b)2 ]

= (1.00)2                                  [ Where a=0.76 and b=0.24 ]

= 1

So, 0.76 x 0.76 + 2 x 0.76 x 0.24 + 0.24 x 0.24 = 1.

(iv) $\frac{7.83*7.83-1.17*1.17}{6.66}$

Solution:

As we have,

$\frac{7.83*7.83-1.17*1.17}{6.66}$

$=\frac{(7.83+1.17)(7.83-1.17)}{6.66}\;\;\;\;\;\;\;[because (a-b)^{2}=(a+b)(a-b)]$

$=\frac{(9.00)(6.66)}{(6.66)}$

= 9

$Therefore, \frac{7.83*7.83-1.17*1.17}{6.66}=9$

4) $If\;x+\frac{1}{x}=11,\;find\;the\;value\;of\;x^{2}+\frac{1}{x^{2}}$

Solution:

As we have,, $x+\frac{1}{x}=11$

So, $(x+\frac{1}{x})^{2}=x^{2}+(\frac{1}{x})^{2}+2*x*\frac{1}{x}$

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

$\Rightarrow (11)^{2}=x^{2}+\frac{1}{x^{2}}+2\;\;\;\;[because x+\frac{1}{x}=11]$

$\Rightarrow 121=x^{2}+\frac{1}{x^{2}}+2$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=119$

5) $If\;x-\frac{1}{x}=-1,\;find\;the\;value\;of\;x^{2}+\frac{1}{x^{2}}$

Solution:

As we have,, $x-\frac{1}{x}=-1$

So, $(x-\frac{1}{x})^{2}=x^{2}+(\frac{1}{x})^{2}-2*x*\frac{1}{x}$

$\Rightarrow (x-\frac{1}{x})^{2}=x^{2}+\frac{1}{x^{2}}-2$

$\Rightarrow (-1)^{2}=x^{2}+\frac{1}{x^{2}}-2\;\;\;\;[because x-\frac{1}{x}=-1]$

$\Rightarrow 2+1=x^{2}+\frac{1}{x^{2}}$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=3$

6) $If\;x+\frac{1}{x}=\sqrt{5},\;find\;the\;value\;of\;x^{2}+\frac{1}{x^{2}}\;and\;x^{4}+\frac{1}{x^{4}}$

Solution:

As we have,,

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

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

$\Rightarrow (\sqrt{5})^{2}=x^{2}+\frac{1}{x^{2}}+2\;\;\;\;[because x+\frac{1}{x}=\sqrt{5}]$

$\Rightarrow 5=x^{2}+\frac{1}{x^{2}}+2$

$\Rightarrow x^{2}+\frac{1}{x^{2}}=3$ …..(1)

So, $(x^{2}+\frac{1}{x^{2}})^{2}=x^{4}+\frac{1}{x^{4}}+2*x^{2}*\frac{1}{x^{2}}$

$\Rightarrow (x^{2}+\frac{1}{x^{2}})^{2}=x^{4}+\frac{1}{x^{4}}+2$

$\Rightarrow 9=x^{4}+\frac{1}{x^{4}}+2\;\;\;\;\;\;\;\;[∵ x^{2}+\frac{1}{x^{2}}=3]$

$\Rightarrow x^{4}+\frac{1}{x^{4}}=7$

Therefore, , $x^{2}+\frac{1}{x^{2}}=3;\;x^{4}+\frac{1}{x^{4}}=7.$

7) $If\;x^{2}+\frac{1}{x^{2}}=66,\;find\;the\;value\;of\;x-\frac{1}{x}$

Solution:

As we have,

$(x-\frac{1}{x})^{2}=x^{2}+(\frac{1}{x})^{2}-2*x*\frac{1}{x}$

$\Rightarrow (x-\frac{1}{x})^{2}=x^{2}+\frac{1}{x^{2}}-2$

$\Rightarrow (x-\frac{1}{x})^{2}=66-2\;\;\;\;[because x^{2}+\frac{1}{x^{2}}=66]$

$\Rightarrow (x-\frac{1}{x})^{2}=64$

$\Rightarrow (x-\frac{1}{x})^{2}=(\pm 8)^{2}$

$\Rightarrow x-\frac{1}{x}=\pm 8$

8) $If\;x^{2}+\frac{1}{x^{2}}=79,\;find\;the\;value\;of\;x+\frac{1}{x}$

Solution:

As we have,

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

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

$\Rightarrow (x+\frac{1}{x})^{2}=79+2\;\;\;\;[because x^{2}+\frac{1}{x^{2}}=79]$

$\Rightarrow (x+\frac{1}{x})^{2}=81$

$\Rightarrow (x+\frac{1}{x})^{2}=(\pm 9)^{2}$

$\Rightarrow x+\frac{1}{x}=\pm 9$

9) If 9x2 + 25y2 = 181 and xy = -6, find the value of 3x + 5y.

Solution:

As we have,

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

=>(3x + 5y)2 = 9x2 + 25y2 + 30xy

= 181 + 30(-6)                 [because, 9x2 + 25y2 = 181 and xy = -6 ]

=>(3x+5y)2 = 1

=>$(3x+5y)^{2}=(\pm 1)^{2}$

=>$3x+5y=\pm 1$

10) If 2x + 3y = 8 and xy = 2, find the value of 4x2 + 9y2.

Solution:

As we have,

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

=> (2x + 3y)2 = 4x2 + 9y2 + 12xy                   [because, 2x + 3y = 8 and xy = 24 ]

=> (8)2 = 4x2 + 9y2 + 24

=> 64 – 24 = 4x2 + 9y2

=> 4x2 + 9y2 = 40

11) If 3x – 7y = 10 and xy = -1, find the value of 9x2 + 49y2.

Solution:

As we have,

(2 – 7y)2 = (3x)2 + (-7y)2 – 2*3x*7y

=> (3x – 7y)2 = 9x2 + 49y2 – 42xy                   [because, 3x – 7y = 10 and xy = -1 ]

=> (10)2 = 9x2 + 49y2 + 42

=> 100 – 42 = 9x2 + 49y2

=> 9x2 + 49y2 = 58

12) Simplify each of the following products:

(i) $(\frac{1}{2}a-3b)(3b+\frac{1}{2}a)(\frac{1}{4}a^{2}+9b^{2})$

Solution: As we have,

$(\frac{1}{2}a-3b)(3b+\frac{1}{2}a)(\frac{1}{4}a^{2}+9b^{2})$

$\Rightarrow [(\frac{1}{2}a)^{2}-(3b)^{2})]\;[(\frac{1}{4}a^{2}+9b^{2})]\;\;\;\;\;\;\;\;\;\;\;[∵ (a+b)(a-b)=a^{2}-b^{2})]$

$\Rightarrow [\frac{1}{4}a^{2}-9b^{2})]\;[\frac{1}{4}a^{2}+9b^{2}]\;\;\;\;\;\;\;\;\;\;\;[∵ (ab)^{2}=a^{2}b^{2})]$

$=[(\frac{1}{4}a^{2})^{2}-(9b^{2})^{2}]\;\;\;\;\;\;\;\;\;\;\;[because (a+b)(a-b)=a^{2}-b^{2})]$

$=\frac{1}{16}a^{4}-81b^{4}$

$Therefore, (\frac{1}{2}a-3b)(3b+\frac{1}{2}a)(\frac{1}{4}a^{2}+9b^{2})=\frac{1}{16}a^{4}-81b^{4}$

(ii) $(m+\frac{n}{7})^{3}\;(m-\frac{n}{7})$

Solution:

As we have,

$(m+\frac{n}{7})^{3}\;(m-\frac{n}{7})$

$=(m+\frac{n}{7})\;(m+\frac{n}{7})\;(m+\frac{n}{7})\;(m-\frac{n}{7})$

$=(m+\frac{n}{7})^{2}\;(m^{2}-(\frac{n}{7})^{2})\;\;\;\;\;\;\;[because (a+b)(a+b)=(a+b)^{2}\;and\;(a+b)(a-b)=a^{2}-b^{2}]$

$=(m+\frac{n}{7})^{2}\;(m^{2}-\frac{n^{2}}{49})$

$Therefore, (m+\frac{n}{7})^{3}\;(m-\frac{n}{7})=(m+\frac{n}{7})^{2}\;(m^{2}-\frac{n^{2}}{49})$

(iii) $(\frac{x}{2}-\frac{2}{5})\;(\frac{2}{5}-\frac{x}{2})-x^{2}+2x$

Solution:

As we have,

$(\frac{x}{2}-\frac{2}{5})\;(\frac{2}{5}-\frac{x}{2})-x^{2}+2x$

$\Rightarrow -(\frac{2}{5}-\frac{x}{2})\;(\frac{2}{5}-\frac{x}{2})-x^{2}+2x$

$\Rightarrow -(\frac{2}{5}-\frac{x}{2})^{2}-x^{2}+2x\;\;\;\;\;\;\;\;[because (a-b)(a-b)=(a-b)^{2}]$

$\Rightarrow -[(\frac{2}{5})^{2}+(\frac{x}{2})^{2}-2(\frac{2}{5})(\frac{x}{2})]-x^{2}+2x$

$\Rightarrow -(\frac{4}{25}+\frac{x^{2}}{4}-\frac{2x}{5})-x^{2}+2x$

$\Rightarrow -\frac{x^{2}}{4}-x^{2}+\frac{2x}{5}+2x-\frac{4}{25}$

$\Rightarrow -\frac{5x^{2}}{4}+\frac{12x}{5}-\frac{4}{25}$

$Therefore, (\frac{x}{2}-\frac{2}{5})\;(\frac{2}{5}-\frac{x}{2})-x^{2}+2x= -\frac{5x^{2}}{4}+\frac{12x}{5}-\frac{4}{25}$

(iv) $(x^{2}+x-2)\;(x^{2}-x+2)$

Solution: As we have,

$(x^{2}+x-2)\;(x^{2}-x+2)$

$[(x)^{2}+(x-2)]\;[(x^{2}-(x+2)]$

=>$(x^{2})^{2}-(x-2) ^{2}$                    [ (a – b) (a + b) = a2 – b2]

=>$x^{4}-(x^{2}+4-4x)\;\;\;\;\;\;\;[because (a-b)^{2}=a^{2}+b^{2}-2ab]$

=>$x^{4}-x^{2}+4x-4$

$Therefore, (x^{2}+x-2)\;(x^{2}-x+2)=x^{4}-x^{2}+4x-4$

(v) $(x^{3}-3x-x)\;(x^{2}-3x+1)$

Solution:

As we have,

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

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

=>$x[(x^{2}-3x)^{2}-(1)^{2}]\;\;\;\;\;\;\;\;\;[because (a+b)(a-b)=a^{2}-b^{2}]$

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

=>$x[x^{4}+9x^{2}-6x^{3}-1]$

=>$x^{5}-6x^{4}+9x^{3}-x$

$Therefore, (x^{3}-3x-x)\;(x^{2}-3x+1)=x^{5}-6x^{4}+9x^{3}-x$

(vi) $(2x^{4}-4x^{2}+1)\;(2x^{4}-4x^{2}-1)$

Solution:

As we have,

$(2x^{4}-4x^{2}+1)\;(2x^{4}-4x^{2}-1)$

=>$[(2x^{4}-4x^{2})^{2}-(1)^{2}]\;\;\;\;\;\;\;[because (a+b)(a-b)=a^{2}-b^{2}]$

=>$[(2x^{4})^{2}+(4x^{2})^{2}-2(2x^{4})(4x^{2})-1]$

=>$4x^{8}-16x^{6}+16x^{4}-1\;\;\;\;\;\;[∵ (a-b)^{2}=a^{2}+b^{2}-2ab]$

$Therefore, (2x^{4}-4x^{2}+1)\;(2x^{4}-4x^{2}-1)=4x^{8}-16x^{6}+16x^{4}-1$

13) Prove that $a^{2}+b^{2}+c^{2}-ab-bc-ca$ is always non-negative for all values of a, b and c.

Solution:

As we have,

$a^{2}+b^{2}+c^{2}-ab-bc-ca$

Multiply and divide by ‘2’

= $\frac{2}{2}\;[a^{2}+b^{2}+c^{2}-ab-bc-ca]$

= $\frac{1}{2}\;[2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca]$

= $\frac{1}{2}\;[a^{2}+a^{2}+b^{2}+b^{2}+c^{2}+c^{2}-2ab-2bc-2ca]$

= $\frac{1}{2}\;[(a^{2}+b^{2}-2ab)+(a^{2}+c^{2}-2ca)+(b^{2}+c^{2}-2bc)]$

= $\frac{1}{2}\;[(a-b)^{2}+(b-c)^{2}+(c-a)^{2}]\;\;\;\;\;\;[because (a-b)^{2}=a^{2}+b^{2}-2ab]$

= $\frac{(a-b)^{2}+(b-c)^{2}+(c-a)^{2}}{2}\geq 0$

$Therefore, a^{2}+b^{2}+c^{2}-ab-bc-ca\geq 0$

Therefore, , $a^{2}+b^{2}+c^{2}-ab-bc-ca\geq 0$ is always non-negative for all values of a, b and c.