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NCERT Solutions

Exploring Algebraic Identities

CBSE · Class 9 · Mathematics

NCERT Solutions for Exploring Algebraic Identities — CBSE Class 9 Mathematics.

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128 Questions Solved · 6 Sections

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EXERCISE SET 4.1

1(i)(7x+4y)2(7x + 4y)^2Show solution
Using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=7xa=7x and b=4yb=4y:

(7x+4y)2=(7x)2+2(7x)(4y)+(4y)2(7x+4y)^2=(7x)^2+2(7x)(4y)+(4y)^2
=49x2+56xy+16y2=49x^2+56xy+16y^2

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1(ii)(75x+32y)2\left(\frac{7}{5}x + \frac{3}{2}y\right)^2Show solution
Using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=75xa=\frac{7}{5}x and b=32yb=\frac{3}{2}y:

(75x+32y)2=(75x)2+2(75x)(32y)+(32y)2\left(\frac{7}{5}x+\frac{3}{2}y\right)^2=\left(\frac{7}{5}x\right)^2+2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right)+\left(\frac{3}{2}y\right)^2
=4925x2+215xy+94y2=\frac{49}{25}x^2+\frac{21}{5}xy+\frac{9}{4}y^2

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1(iii)(2.5p+1.5q)2(2.5p + 1.5q)^2Show solution
Using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=2.5pa=2.5p and b=1.5qb=1.5q:

(2.5p+1.5q)2=(2.5p)2+2(2.5p)(1.5q)+(1.5q)2(2.5p+1.5q)^2=(2.5p)^2+2(2.5p)(1.5q)+(1.5q)^2
=6.25p2+7.5pq+2.25q2=6.25p^2+7.5pq+2.25q^2

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1(iv)(34s+8t)2\left(\frac{3}{4}s + 8t\right)^2Show solution
Using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=34sa=\frac{3}{4}s and b=8tb=8t:

(34s+8t)2=(34s)2+2(34s)(8t)+(8t)2\left(\frac{3}{4}s+8t\right)^2=\left(\frac{3}{4}s\right)^2+2\left(\frac{3}{4}s\right)(8t)+(8t)^2
=916s2+12st+64t2=\frac{9}{16}s^2+12st+64t^2

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1(v)(x+12y)2\left(x + \frac{1}{2y}\right)^2Show solution
Using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=xa=x and b=12yb=\frac{1}{2y}:

(x+12y)2=x2+2⋅x⋅12y+(12y)2\left(x+\frac{1}{2y}\right)^2=x^2+2\cdot x\cdot \frac{1}{2y}+\left(\frac{1}{2y}\right)^2
=x2+xy+14y2=x^2+\frac{x}{y}+\frac{1}{4y^2}

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1(vi)(1x+1y)2\left(\frac{1}{x} + \frac{1}{y}\right)^2Show solution
Using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=1xa=\frac1x and b=1yb=\frac1y:

(1x+1y)2=(1x)2+2(1x)(1y)+(1y)2\left(\frac1x+\frac1y\right)^2=\left(\frac1x\right)^2+2\left(\frac1x\right)\left(\frac1y\right)+\left(\frac1y\right)^2
=1x2+2xy+1y2=\frac{1}{x^2}+\frac{2}{xy}+\frac{1}{y^2}

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2(i)(64)2(64)^2Show solution
64264^2 can be found directly:

642=64×64=409664^2=64\times 64=4096

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2(ii)(105)2(105)^2Show solution
Use (a+b)2(a+b)^2 with 105=100+5105=100+5:

(105)2=(100+5)2=1002+2(100)(5)+52(105)^2=(100+5)^2=100^2+2(100)(5)+5^2
=10000+1000+25=11025=10000+1000+25=11025

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2(iii)(205)2(205)^2Show solution
Use (a+b)2(a+b)^2 with 205=200+5205=200+5:

(205)2=(200+5)2=2002+2(200)(5)+52(205)^2=(200+5)^2=200^2+2(200)(5)+5^2
=40000+2000+25=42025=40000+2000+25=42025

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1(i)Using the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, expand the following:

(7x+4y)2(7x + 4y)^2Show solution
Using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=7xa=7x and b=4yb=4y:

(7x+4y)2=(7x)2+2(7x)(4y)+(4y)2(7x+4y)^2=(7x)^2+2(7x)(4y)+(4y)^2
=49x2+56xy+16y2=49x^2+56xy+16y^2

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1(ii)Using the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, expand the following:

(75x+32y)2\left(\frac{7}{5}x + \frac{3}{2}y\right)^2Show solution
Using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=75xa=\frac{7}{5}x and b=32yb=\frac{3}{2}y:

(75x+32y)2=(75x)2+2(75x)(32y)+(32y)2\left(\frac{7}{5}x+\frac{3}{2}y\right)^2=\left(\frac{7}{5}x\right)^2+2\left(\frac{7}{5}x\right)\left(\frac{3}{2}y\right)+\left(\frac{3}{2}y\right)^2
=4925x2+215xy+94y2=\frac{49}{25}x^2+\frac{21}{5}xy+\frac{9}{4}y^2

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1(iii)Using the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, expand the following:

(2.5p+1.5q)2(2.5p + 1.5q)^2Show solution
Using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=2.5pa=2.5p and b=1.5qb=1.5q:

(2.5p+1.5q)2=(2.5p)2+2(2.5p)(1.5q)+(1.5q)2(2.5p+1.5q)^2=(2.5p)^2+2(2.5p)(1.5q)+(1.5q)^2
=6.25p2+7.5pq+2.25q2=6.25p^2+7.5pq+2.25q^2

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1(iv)(34s+8t)2(\frac{3}{4}s + 8t)^2Show solution
Using (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=34sa=\frac{3}{4}s and b=8tb=8t:

(34s+8t)2=(34s)2+2(34s)(8t)+(8t)2\left(\frac{3}{4}s+8t\right)^2=\left(\frac{3}{4}s\right)^2+2\left(\frac{3}{4}s\right)(8t)+(8t)^2
=916s2+12st+64t2=\frac{9}{16}s^2+12st+64t^2

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2(i)Using the same identity, find the values of the following:

(64)2(64)^2Show solution
642=64×64=409664^2=64\times 64=4096.

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2(ii)Using the same identity, find the values of the following:

(105)2(105)^2Show solution
1052=(100+5)2=1002+2⋅100⋅5+52=10000+1000+25=11025105^2=(100+5)^2=100^2+2\cdot100\cdot5+5^2=10000+1000+25=11025.

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2(iii)Using the same identity, find the values of the following:

(205)2(205)^2Show solution
2052=(200+5)2=2002+2⋅200⋅5+52=40000+2000+25=42025205^2=(200+5)^2=200^2+2\cdot200\cdot5+5^2=40000+2000+25=42025.

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1(i)Using the identity (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2, expand the following:Show solution
The expansions are:

- (7x+4y)2=49x2+56xy+16y2(7x+4y)^2=49x^2+56xy+16y^2
- (75x+32y)2=4925x2+215xy+94y2\left(\frac{7}{5}x+\frac{3}{2}y\right)^2=\frac{49}{25}x^2+\frac{21}{5}xy+\frac{9}{4}y^2
- (2.5p+1.5q)2=6.25p2+7.5pq+2.25q2(2.5p+1.5q)^2=6.25p^2+7.5pq+2.25q^2
- (34s+8t)2=916s2+12st+64t2\left(\frac{3}{4}s+8t\right)^2=\frac{9}{16}s^2+12st+64t^2
- (x+12y)2=x2+xy+14y2\left(x+\frac{1}{2y}\right)^2=x^2+\frac{x}{y}+\frac{1}{4y^2}
- (1x+1y)2=1x2+2xy+1y2\left(\frac{1}{x}+\frac{1}{y}\right)^2=\frac{1}{x^2}+\frac{2}{xy}+\frac{1}{y^2}

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2(i)Using the same identity, find the values of the following:Show solution
The values are:

- 642=409664^2=4096
- 1052=11025105^2=11025
- 2052=42025205^2=42025

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EXERCISE SET 4.2

1(i)9x2+24xy+16y29x^2 + 24xy + 16y^2Show solution

9x2+24xy+16y2=(3x)2+2(3x)(4y)+(4y)29x^2+24xy+16y^2=(3x)^2+2(3x)(4y)+(4y)^2

So it matches the identity (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2 with a=3xa=3x and b=4yb=4y, hence

9x2+24xy+16y2=(3x+4y)29x^2+24xy+16y^2=(3x+4y)^2.

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1(ii)4s2+20st+25t24s^2 + 20st + 25t^2Show solution

4s2+20st+25t2=(2s)2+2(2s)(5t)+(5t)24s^2+20st+25t^2=(2s)^2+2(2s)(5t)+(5t)^2

So,

4s2+20st+25t2=(2s+5t)24s^2+20st+25t^2=(2s+5t)^2.

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1(iii)49x2+28xy+4y249x^{2} + 28xy + 4y^{2}Show solution

49x2+28xy+4y2=(7x)2+2(7x)(2y)+(2y)249x^2+28xy+4y^2=(7x)^2+2(7x)(2y)+(2y)^2

Therefore,

49x2+28xy+4y2=(7x+2y)249x^2+28xy+4y^2=(7x+2y)^2.

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1(iv)64p2+323pq+49q264p^{2} + \frac{32}{3}pq + \frac{4}{9}q^{2}Show solution

64p2+323pq+49q2=(8p)2+2(8p)(23q)+(23q)264p^2+\frac{32}{3}pq+\frac{4}{9}q^2=(8p)^2+2(8p)\left(\frac23 q\right)+\left(\frac23 q\right)^2

So,

64p2+323pq+49q2=(8p+23q)264p^2+\frac{32}{3}pq+\frac{4}{9}q^2=(8p+\frac23 q)^2.

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1(v)3a2+4ab+43b23a^{2} + 4ab + \frac{4}{3}b^{2}Show solution

First take out the common factor 3:

3a2+4ab+43b2=3(a2+43ab+49b2)3a^2+4ab+\frac{4}{3}b^2=3\left(a^2+\frac{4}{3}ab+\frac{4}{9}b^2\right)

Now compare inside the brackets with (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2:

a2+43ab+49b2=(a+23b)2a^2+\frac{4}{3}ab+\frac{4}{9}b^2=\left(a+\frac{2}{3}b\right)^2

So the factorisation is

3(a+23b)23\left(a+\frac{2}{3}b\right)^2

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1(vi)95s2+6sv+5v2\frac{9}{5}s^{2} + 6sv + 5v^{2}Show solution

Take out 15\frac{1}{5}:

95s2+6sv+5v2=15(9s2+30sv+25v2)\frac{9}{5}s^2+6sv+5v^2=\frac{1}{5}(9s^2+30sv+25v^2)

Now,

9s2+30sv+25v2=(3s+5v)29s^2+30sv+25v^2=(3s+5v)^2

Therefore,

95s2+6sv+5v2=15(3s+5v)2\frac{9}{5}s^2+6sv+5v^2=\frac{1}{5}(3s+5v)^2

Equivalent factor form: 15(3s+5v)2\frac{1}{5}(3s+5v)^2.

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2(i)(79)2(79)^{2}Show solution
Use (a−b)2(a-b)^2 with 79=80−179=80-1:

(79)2=(80−1)2=802−2⋅80⋅1+12(79)^2=(80-1)^2=80^2-2\cdot80\cdot1+1^2
=6400−160+1=6241=6400-160+1=6241

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2(ii)(193)2(193)^{2}Show solution
Use the identity (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2 with a=200a=200 and b=7b=7:

(193)2=(200−7)2=2002−2(200)(7)+72=40000−2800+49=37249(193)^2=(200-7)^2=200^2-2(200)(7)+7^2=40000-2800+49=37249.

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2(iii)(299)2(299)^{2}Show solution
Use the identity (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2 with a=300a=300 and b=1b=1:

(299)2=(300−1)2=3002−2(300)(1)+12=90000−600+1=89401(299)^2=(300-1)^2=300^2-2(300)(1)+1^2=90000-600+1=89401.

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1(i)Factor completely:Show solution
Factor each expression by matching it to a perfect square identity.

- 9x2+24xy+16y2=(3x)2+2(3x)(4y)+(4y)2=(3x+4y)29x^2+24xy+16y^2=(3x)^2+2(3x)(4y)+(4y)^2=(3x+4y)^2
- 4s2+20st+25t2=(2s)2+2(2s)(5t)+(5t)2=(2s+5t)24s^2+20st+25t^2=(2s)^2+2(2s)(5t)+(5t)^2=(2s+5t)^2
- 49x2+28xy+4y2=(7x)2+2(7x)(2y)+(2y)2=(7x+2y)249x^2+28xy+4y^2=(7x)^2+2(7x)(2y)+(2y)^2=(7x+2y)^2
- 64p2+323pq+49q2=(8p)2+2(8p)(2q3)+(2q3)2=(8p+2q3)264p^2+\frac{32}{3}pq+\frac{4}{9}q^2=(8p)^2+2(8p)\left(\frac{2q}{3}\right)+\left(\frac{2q}{3}\right)^2=(8p+\frac{2q}{3})^2

For the last two, as printed they do not match a direct square pattern unless a common factor is taken first; the textbook hint says to look for such a factor. Without altering the expressions, they are not direct perfect squares in the same way as the others.

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2(i)Find the values of the following using the identity

(a−b)2=a2−2ab+b2(a - b)^{2} = a^{2} - 2ab + b^{2}.Show solution
Use (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2.

- 792=(80−1)2=802−2(80)(1)+12=6400−160+1=624179^2=(80-1)^2=80^2-2(80)(1)+1^2=6400-160+1=6241
- 1932=(200−7)2=2002−2(200)(7)+72=40000−2800+49=37249193^2=(200-7)^2=200^2-2(200)(7)+7^2=40000-2800+49=37249
- 2992=(300−1)2=3002−2(300)(1)+12=90000−600+1=89401299^2=(300-1)^2=300^2-2(300)(1)+1^2=90000-600+1=89401

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EXERCISE SET 4.3

1(i)1172117^2Show solution
Use (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2:

1172=(100+17)2=1002+2(100)(17)+172=10000+3400+289=13689117^2=(100+17)^2=100^2+2(100)(17)+17^2=10000+3400+289=13689.

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1(ii)78278^2Show solution
Use (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2:

782=(80−2)2=802−2(80)(2)+22=6400−320+4=608478^2=(80-2)^2=80^2-2(80)(2)+2^2=6400-320+4=6084.

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1(iii)1982198^2Show solution
Use (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2:

1982=(200−2)2=2002−2(200)(2)+22=40000−800+4=39204198^2=(200-2)^2=200^2-2(200)(2)+2^2=40000-800+4=39204.

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1(iv)2142214^2Show solution
Use (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2:

2142=(200+14)2=2002+2(200)(14)+142=40000+5600+196=45796214^2=(200+14)^2=200^2+2(200)(14)+14^2=40000+5600+196=45796.

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1(v)110421104^2Show solution
Use (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca:

11042=(1100+4)2=11002+2(1100)(4)+42=1210000+8800+16=12188161104^2=(1100+4)^2=1100^2+2(1100)(4)+4^2=1210000+8800+16=1218816

But checking directly with (1100+4)2(1100+4)^2 gives:

11002=12100001100^2=1210000
2⋅1100⋅4=88002\cdot1100\cdot4=8800
42=164^2=16

So the correct value is 1218816. The computed answer is not among the listed options because this is a calculation, not an option-based question.

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1(vi)112021120^2Show solution
Use (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2:

11202=(1100+20)2=11002+2(1100)(20)+2021120^2=(1100+20)^2=1100^2+2(1100)(20)+20^2
=1210000+44000+400=1254400=1210000+44000+400=1254400.

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2(i)16y2−24y+916y^2 - 24y + 9Show solution
Compare 16y2−24y+916y^2-24y+9 with a2−2ab+b2a^2-2ab+b^2.

Here 16y2=(4y)216y^2=(4y)^2, 9=329=3^2, and −24y=−2(4y)(3)-24y=-2(4y)(3).

So, 16y2−24y+9=(4y−3)216y^2-24y+9=(4y-3)^2.

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2(ii)94s2+6st+4t2\frac{9}{4}s^2 + 6st + 4t^2Show solution
Compare 94s2+6st+4t2\frac{9}{4}s^2+6st+4t^2 with a2+2ab+b2a^2+2ab+b^2.

Take a=32sa=\frac{3}{2}s and b=2tb=2t.
Then
- a2=94s2a^2=\frac{9}{4}s^2
- 2ab=2(32s)(2t)=6st2ab=2\left(\frac{3}{2}s\right)(2t)=6st
- b2=4t2b^2=4t^2

So, 94s2+6st+4t2=(32s+2t)2\frac{9}{4}s^2+6st+4t^2=\left(\frac{3}{2}s+2t\right)^2.

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2(iii)m29+mk3+k24+3nk+2mn+9n2\frac{m^2}{9} + \frac{mk}{3} + \frac{k^2}{4} + 3nk + 2mn + 9n^2Show solution
Group the terms:

m29+mk3+k24+2mn+3nk+9n2\frac{m^2}{9}+\frac{mk}{3}+\frac{k^2}{4}+2mn+3nk+9n^2

This matches the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca with

a=m3,  b=k2,  c=3na=\frac{m}{3},\; b=\frac{k}{2},\; c=3n.

Then
- a2=m29a^2=\frac{m^2}{9}
- b2=k24b^2=\frac{k^2}{4}
- c2=9n2c^2=9n^2
- 2ab=mk32ab=\frac{mk}{3}
- 2bc=3nk2bc=3nk
- 2ca=2mn2ca=2mn

So the factorised form is (m3+k2+3n)2(\frac{m}{3}+\frac{k}{2}+3n)^2.

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2(iv)p216−2+16p2\frac{p^2}{16} - 2 + \frac{16}{p^2}Show solution
Compare p216−2+16p2\frac{p^2}{16}-2+\frac{16}{p^2} with a2−2ab+b2a^2-2ab+b^2.

Take a=p4a=\frac{p}{4} and b=4pb=\frac{4}{p}.
Then
- a2=p216a^2=\frac{p^2}{16}
- b2=16p2b^2=\frac{16}{p^2}
- −2ab=−2(p4)(4p)=−2-2ab=-2\left(\frac{p}{4}\right)\left(\frac{4}{p}\right)=-2

So, p216−2+16p2=(p4−4p)2\frac{p^2}{16}-2+\frac{16}{p^2}=\left(\frac{p}{4}-\frac{4}{p}\right)^2.

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2(v)9a2+4b2+c2−12ab+6ac−4bc9a^2 + 4b^2 + c^2 - 12ab + 6ac - 4bcShow solution
Compare 9a2+4b2+c2−12ab+6ac−4bc9a^2+4b^2+c^2-12ab+6ac-4bc with (x+y+z)2=x2+y2+z2+2xy+2yz+2zx(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2zx.

Take x=3ax=3a, y=−2by=-2b, z=cz=c.
Then
- x2=9a2x^2=9a^2
- y2=4b2y^2=4b^2
- z2=c2z^2=c^2
- 2xy=2(3a)(−2b)=−12ab2xy=2(3a)(-2b)=-12ab
- 2yz=2(−2b)(c)=−4bc2yz=2(-2b)(c)=-4bc
- 2zx=2(c)(3a)=6ac2zx=2(c)(3a)=6ac

So the factorisation is (3a−2b+c)2(3a-2b+c)^2.

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3(i)(p+3q+7r)2(p + 3q + 7r)^2Show solution
Use (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca with a=pa=p, b=3qb=3q, c=7rc=7r.

So,

(p+3q+7r)2=p2+(3q)2+(7r)2+2(p)(3q)+2(3q)(7r)+2(7r)(p)(p+3q+7r)^2=p^2+(3q)^2+(7r)^2+2(p)(3q)+2(3q)(7r)+2(7r)(p)

=p2+9q2+49r2+6pq+42qr+14pr=p^2+9q^2+49r^2+6pq+42qr+14pr.

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3(ii)(3x−2y+4z)2(3x - 2y + 4z)^2Show solution
Use (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca with a=3xa=3x, b=−2yb=-2y, c=4zc=4z.

Then

(3x−2y+4z)2=(3x)2+(−2y)2+(4z)2+2(3x)(−2y)+2(−2y)(4z)+2(4z)(3x)(3x-2y+4z)^2=(3x)^2+(-2y)^2+(4z)^2+2(3x)(-2y)+2(-2y)(4z)+2(4z)(3x)

=9x2+4y2+16z2−12xy−16yz+24xz=9x^2+4y^2+16z^2-12xy-16yz+24xz.

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4Is this an identity?

(a+b−c)2+(a−b+c)2+(a−b−c)2=2a2+2b2+2c2.(a + b - c)^2 + (a - b + c)^2 + (a - b - c)^2 = 2a^2 + 2b^2 + 2c^2.Show solution
Expand each square and add:

(a+b−c)2=a2+b2+c2+2ab−2ac−2bc(a+b-c)^2=a^2+b^2+c^2+2ab-2ac-2bc

(a−b+c)2=a2+b2+c2−2ab+2ac−2bc(a-b+c)^2=a^2+b^2+c^2-2ab+2ac-2bc

(a−b−c)2=a2+b2+c2−2ab−2ac+2bc(a-b-c)^2=a^2+b^2+c^2-2ab-2ac+2bc

Adding them gives:

3a2+3b2+3c23a^2+3b^2+3c^2

So the left side equals 2a2+2b2+2c22a^2+2b^2+2c^2 only if the statement is adjusted? Wait, the sum actually is 3a2+3b2+3c23a^2+3b^2+3c^2, not 2a2+2b2+2c22a^2+2b^2+2c^2.

Hence the given statement is not an identity.

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1(i)35235^2Show solution
Use (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2:

352=(30+5)235^2=(30+5)^2 or (40−5)2(40-5)^2.
Using (30+5)2(30+5)^2:

352=302+2(30)(5)+52=900+300+25=122535^2=30^2+2(30)(5)+5^2=900+300+25=1225.

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1(ii)65265^2Show solution
Use (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2:

652=(60+5)2=602+2(60)(5)+52=3600+600+25=422565^2=(60+5)^2=60^2+2(60)(5)+5^2=3600+600+25=4225.

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1(iii)85285^2Show solution
Use (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2:

852=(80+5)2=802+2(80)(5)+52=6400+800+25=722585^2=(80+5)^2=80^2+2(80)(5)+5^2=6400+800+25=7225.

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1(iv)1052105^2Show solution
Use (a+b)2=a2+2ab+b2(a+b)^2=a^2+2ab+b^2:

1052=(100+5)2=1002+2(100)(5)+52=10000+1000+25=11025105^2=(100+5)^2=100^2+2(100)(5)+5^2=10000+1000+25=11025.

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2Observe the two rows of figures below. They represent an algebraic identity. Try to identify it.Show solution
The two rows of figures represent the expansion of a square of side a+b+ca+b+c into smaller squares and rectangles. So the identity is

(a+b+c)2=a2+b2+c2+2ab+2bc+2ca.(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca.

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4Is this an identity?Show solution
Expand the left side:

(a+b−c)2+(a−b+c)2+(a−b−c)2(a+b-c)^2+(a-b+c)^2+(a-b-c)^2

=(a2+b2+c2+2ab−2ac−2bc)=(a^2+b^2+c^2+2ab-2ac-2bc)
+(a2+b2+c2−2ab+2ac−2bc)+(a^2+b^2+c^2-2ab+2ac-2bc)
+(a2+b2+c2−2ab−2ac+2bc)+(a^2+b^2+c^2-2ab-2ac+2bc)

=3a2+3b2+3c2−2ab−2ac−2bc=3a^2+3b^2+3c^2-2ab-2ac-2bc

So it does not equal 2a2+2b2+2c22a^2+2b^2+2c^2 for all values. Therefore, it is not an identity.

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1(i)Find the following squares using one of the above identities. Determine which of these identities will make these calculations easier.Show solution
Compute each square using the most suitable identity:

- 1172=(100+17)2=10000+3400+289=13689117^2=(100+17)^2=10000+3400+289=13689
- 782=(80−2)2=6400−320+4=608478^2=(80-2)^2=6400-320+4=6084
- 1982=(200−2)2=40000−800+4=39204198^2=(200-2)^2=40000-800+4=39204
- 2142=(200+14)2=40000+5600+196=45796214^2=(200+14)^2=40000+5600+196=45796
- 11042=(1100+4)2=1210000+8800+16=12188161104^2=(1100+4)^2=1210000+8800+16=1218816
- 11202=(1100+20)2=1210000+44000+400=12544001120^2=(1100+20)^2=1210000+44000+400=1254400

For these, the square of a sum or square of a difference identities are easiest, depending on which nearby round number is chosen.

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EXERCISE SET 4.4

1(i)s^2 - 11s + 24 = (\underline{\hspace{2cm}}) (\underline{\hspace{2cm}})Show solution
Find two numbers whose product is 2424 and sum is −11-11. They are −3-3 and −8-8.

So,

s2−11s+24=s2−3s−8s+24=s(s−3)−8(s−3)=(s−3)(s−8).s^2 - 11s + 24 = s^2 - 3s - 8s + 24 = s(s-3) - 8(s-3) = (s-3)(s-8).

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1(ii)(\underline{\hspace{2cm}})(x + 1) = (3x^2 - 4x - 7)Show solution
Let the missing factor be AA. Then

A(x+1)=3x2−4x−7.A(x+1)=3x^2-4x-7.

Factor by grouping:

3x2−4x−7=3x2+3x−7x−7=3x(x+1)−7(x+1)=(3x−7)(x+1).3x^2-4x-7=3x^2+3x-7x-7=3x(x+1)-7(x+1)=(3x-7)(x+1).

So the missing factor is 3x−73x-7.

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1(iii)10x^2 - 11x - 6 = (2x - \underline{\hspace{2cm}}) (\underline{\hspace{2cm}} + 2)Show solution
We need

(2x−_)(_+2)=10x2−11x−6.(2x-\_)(\_+2)=10x^2-11x-6.

Try splitting the middle term by comparing with (2x−a)(5x+2)(2x-a)(5x+2):

(2x−a)(5x+2)=10x2+4x−5ax−2a=10x2+(4−5a)x−2a. (2x-a)(5x+2)=10x^2+4x-5ax-2a = 10x^2+(4-5a)x-2a.

Match coefficients with 10x2−11x−610x^2-11x-6:

−2a=−6⇒a=3,-2a=-6 \Rightarrow a=3,

and then

4−5(3)=4−15=−11.4-5(3)=4-15=-11.

So,

10x2−11x−6=(2x−3)(5x+2).10x^2-11x-6=(2x-3)(5x+2).

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1(iv)6x^2 + 7x + 2 = (\underline{\hspace{2cm}}) (\underline{\hspace{2cm}})Show solution
Factor 6x2+7x+26x^2+7x+2 by splitting the middle term.

We need two numbers whose product is 6⋅2=126\cdot 2=12 and sum is 77. They are 33 and 44.

6x2+7x+2=6x2+3x+4x+2=3x(2x+1)+2(2x+1)=(3x+2)(2x+1).6x^2+7x+2=6x^2+3x+4x+2=3x(2x+1)+2(2x+1)=(3x+2)(2x+1).

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2(i)(41)2(41)^2Show solution
Use (a+b)2(a+b)^2:

412=(40+1)2=402+2⋅40⋅1+12=1600+80+1=1681.41^2=(40+1)^2=40^2+2\cdot40\cdot1+1^2=1600+80+1=1681.

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2(ii)(27)2(27)^2Show solution
Use (a−b)2(a-b)^2:

272=(30−3)2=302−2⋅30⋅3+32=900−180+9=729.27^2=(30-3)^2=30^2-2\cdot30\cdot3+3^2=900-180+9=729.

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2(iii)(23×17)(23 \times 17)Show solution
Use distributive property:

23×17=23×(20−3)=460−69=391.23\times17=23\times(20-3)=460-69=391.

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2(iv)(135)2(135)^2Show solution
Use (a−b)2(a-b)^2:

1352=(140−5)2=1402−2⋅140⋅5+52=19600−1400+25=18225.135^2=(140-5)^2=140^2-2\cdot140\cdot5+5^2=19600-1400+25=18225.

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2(v)(97)2(97)²Show solution
Use (a−b)2(a-b)^2:

972=(100−3)2=1002−2⋅100⋅3+32=10000−600+9=9409.97^2=(100-3)^2=100^2-2\cdot100\cdot3+3^2=10000-600+9=9409.

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2(vi)(18×29)(18 × 29)Show solution
Use distributive property:

18×29=18×(30−1)=540−18=522.18\times29=18\times(30-1)=540-18=522.

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2(vii)(34×43)(34 × 43)Show solution
Use distributive property:

34×43=34×(40+3)=1360+102=1462.34\times43=34\times(40+3)=1360+102=1462.

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2(viii)(205)2(205)²Show solution
Use (a+b)2(a+b)^2:

2052=(200+5)2=2002+2⋅200⋅5+52=40000+2000+25=42025.205^2=(200+5)^2=200^2+2\cdot200\cdot5+5^2=40000+2000+25=42025.

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3(i)9a² + b² + 4c² - 6ab + 12ac - 4bcShow solution
Use the identity (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2.

Here,

9a2=(3a)2,b2=b2,4c2=(2c)2.9a^2=(3a)^2,\quad b^2=b^2,\quad 4c^2=(2c)^2.

Check the middle terms:

−6ab+12ac−4bc=−2(3a)(b)+2(3a)(2c)−2(b)(2c).-6ab+12ac-4bc = -2(3a)(b)+2(3a)(2c)-2(b)(2c).

So the expression is a perfect square:

9a2+b2+4c2−6ab+12ac−4bc=(3a−b+2c)2.9a^2+b^2+4c^2-6ab+12ac-4bc=(3a-b+2c)^2.

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3(ii)16s² + 25t² - 40stShow solution
Use (a−b)2=a2−2ab+b2(a-b)^2=a^2-2ab+b^2.

16s2=(4s)2,25t2=(5t)2,−40st=−2(4s)(5t).16s^2=(4s)^2,\quad 25t^2=(5t)^2,\quad -40st=-2(4s)(5t).

Therefore,

16s2+25t2−40st=(4s−5t)2.16s^2+25t^2-40st=(4s-5t)^2.

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3(iii)r² - r - 42
3(iv)49g² + 14gh + h²
3(v)64u² + 121v² + 4w² - 176uv - 32uw + 44vw
1(i)Fill in the blanks to complete the following identities:
s2−11s+24=(‾)(‾)s^2 - 11s + 24 = (\underline{\hspace{2cm}}) (\underline{\hspace{2cm}})
1(ii)Fill in the blanks to complete the following identities:
(‾)(x+1)=(3x2−4x−7)(\underline{\hspace{2cm}})(x + 1) = (3x^2 - 4x - 7)
1(iii)Fill in the blanks to complete the following identities:
10x2−11x−6=(2x−‾)(‾+2)10x^2 - 11x - 6 = (2x - \underline{\hspace{2cm}}) (\underline{\hspace{2cm}} + 2)
1(iv)Fill in the blanks to complete the following identities:
6x2+7x+2=(‾)(‾)6x^2 + 7x + 2 = (\underline{\hspace{2cm}}) (\underline{\hspace{2cm}})
1(i)Fill in the blanks to complete the following identities:

EXERCISE SET 4.5

1(i)3p2−3pq−18q2p2+3pq−10q2\frac{3p^2 - 3pq - 18q^2}{p^2 + 3pq - 10q^2}
1(ii)n3−3n2m+3nm2−m35m2−10mn+5n2\frac{n^3 - 3n^2m + 3nm^2 - m^3}{5m^2 - 10mn + 5n^2}
1(iii)\frac{w^3 - v^3 + x^3 + 3wvx}{w^2 + v^2 + x^2 - 2wv - 2vx + 2wx}
1(iv)4y2−20yz+25z2(25z2−4y2)\frac{4y^2 - 20yz + 25z^2}{(25z^2 - 4y^2)}
1(v)(x2+x−6)(x2−7x+12)(x2−6x+8)(x2−9)\frac{(x^2 + x - 6)(x^2 - 7x + 12)}{(x^2 - 6x + 8)(x^2 - 9)}
1(vi)p4−16p2−4p+4\frac{p^4 - 16}{p^2 - 4p + 4}
2Draw Saira's rectangle using these pieces.
3Find the length and breadth of the pool.
4(i)(−3x+4)2(-3x + 4)^2
4(ii)(2s + 7)(2s - 7)
4(iii)(p2+12)(p2−12)\left(p^2 + \frac{1}{2}\right)\left(p^2 - \frac{1}{2}\right)
4(iv)(2n + 7)(2n - 7)
4(v)(s−2t)(s2+2st+4t2)(s - 2t)(s^2 + 2st + 4t^2)
4(vi)(12r−4r)2\left(\frac{1}{2r} - 4r\right)^2
4(vii)(−3m+4k−l)2(-3m + 4k - l)^2
4(viii)(x−13y)3\left(x - \frac{1}{3}y\right)^3
4(ix)(72k−23m)3\left(\frac{7}{2}k - \frac{2}{3}m\right)^3
5(i)17×2117 \times 21
5(ii)104×96104 \times 96
5(iii)24×1624 \times 16
5(iv)1473147^3
5(v)1993199^3
5(vi)1273127^3
5(vii)(−107)3(-107)^3
5(viii)(−299)3(-299)^3
6(i)4x2+4x+14x^2 + 4x + 1
6(ii)9(3a3−24b3)9(3a^3 - 24b^3)
6(iii)s3+125t3s^3 + 125t^3
7The village playground is shaped as a square of side 40 metres. A path of width ss metres is created around the playground for people to walk. Find an expression for the area of the path in terms of ss.
8If a number plus its reciprocal equals 103\frac{10}{3}, find the number.
9A rectangular pool has area 2x2+7x+32x^2 + 7x + 3 square *hastas*. If its width is 2x+12x + 1 *hastas*, find its length. *Hasta* was a unit used to measure length.
10If both x−2x - 2 and x−12x - \frac{1}{2} are factors of px2+5x+rpx^2 + 5x + r, show that p=rp = r.
11If a+b+c=5a + b + c = 5 and ab+bc+ca=10ab + bc + ca = 10, then prove that a3+b3+c3−3abc=−25a^3 + b^3 + c^3 - 3abc = -25.
12By factoring the expression, check that n3−nn^3 - n is always divisible by 6 for all natural numbers nn. Give reasons.
13(i)x^3 + y^3 - 12xy + 64, when x+y=−4x + y = -4
13(ii)x^3 - 8y^3 - 36xy - 216, when x=2y+6x = 2y + 6
1(i)Simplify the following rational expressions assuming that the expressions in the denominators are not equal to zero:
4(i)4x2+4x+14x2−1\frac{4x^2 + 4x + 1}{4x^2 - 1}
4(ii)9(3a3−24b3)9a2−36b2\frac{9(3a^3 - 24b^3)}{9a^2 - 36b^2}
4(iii)s3+125t3s2−2st−35t2\frac{s^3 + 125t^3}{s^2 - 2st - 35t^2}
5(i)25a2−30ab+9b225a^2 - 30ab + 9b^2
5(ii)36s2−49t236s^2 - 49t^2
6(i)6a2−24b26a^2 - 24b^2
6(ii)3ps2−15ps+12p3ps^2 - 15ps + 12p
13(i)x3+y3−12xy+64x^3 + y^3 - 12xy + 64
13(ii)x3−8y3−36xy−216x^3 - 8y^3 - 36xy - 216

Think and Reflect

1What can you say about aa and bb if (a+b)2<a2+b2(a + b)^2 < a^2 + b^2?
2What can you say about aa and bb if (a+b)2>a2+b2(a + b)^2 > a^2 + b^2?
3When will (a+b)2(a + b)^2 be equal to a2+b2a^2 + b^2?
1Figure out the product of x+2x + 2 and x+3x + 3 using algebra tiles.
2Lay out algebra tiles for x2+11x+30x^2 + 11x + 30 in such a way that you will see its factors.
6Suppose 7x7x is split as 2x+5x2x + 5x; can a similar rectangular arrangement be formed? Consider other possibilities and check.
7James and Reshma were talking about algebraic identities they learnt in school.

James: (a - b)² (a + b) = (a² - 2ab + b²)(a + b)

Reshma: I have a different idea. (a - b)² (a + b) = (a - b) [(a - b)(a + b)] = (a - b)(a² - b²)

I will find this product to get the answer.

According to you, who is correct and why?
8Try to combine more such identities and find new results.
9What do you think (a + b)³ will look like?
10Label the squares and rectangles in Fig. 4.4 so that it represents the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca.

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