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USE CODE: SELF10

Chapter 4: Algebraic Identities Exercise – 4.2

Question: 1

Write the following in the expand form:

(i) (a + 2b + c)2

(ii) (2a − 3b − c)2

(iii) (−3x + y + z)2

(iv) (m + 2n − 5p)2

(v) (2 + x − 2y)2

(vi) (a2 + b2 + c2)2

(vii) (ab + bc + ca)2

(viii) (x/y + y/z + z/x)2

(ix) (a/bc + b/ac + c/ab)2

(x) (x + 2y + 4z)2

(xi) (2x − y + z)2

(xii) (−2x + 3y + 2z)2

Solution:

(i) We have,

(a + 2b + c)2 = a2 + (2b)2 + c2 + 2a(2b) + 2ac + 2(2b)c

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

∴ (a + 2b + c)2 = a2 + 4b2 + c2 + 4ab + 2ac + 4bc

(ii) We have,

(2a − 3b − c)2 = [(2a) + (−3b) + (−c)]2

(2a)2 +( −3b)2 + (−c)2 + 2(2a)(−3b) + 2(−3b)(−c) + 2(2a)(−c)

[∴ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

4a2 + 9b2 + c2 − 12ab + 6bc − 4ca

∴ (2a - 3b - c)2 = 4x2 + 9y2 + c2 - 12ab + 6bc - 4ca

(iii) We have,

(−3x + y + z)2 = [(−3x)2 + y2 + z2 + 2(−3x)y + 2yz + 2(−3x)z

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

9x2 + y2 + z2 − 6xy + 2yz − 6xz

(−3x + y + z)2 = 9x2 + y2 + z2 − 6xy + 2xy − 6xy

(iv) We have,

(m + 2n − 5p)2 = m2 + (2n)2 + (−5p)2 + 2m × 2n + (2 × 2n × −5p) + 2m × −5p

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

(m + 2n − 5p)2 = m2 + 4n2 + 25p2 + 4mn − 20np − 10pm

(v) We have,

(2 + x − 2y)2 = 22 + x2 + (−2y)2 + 2(2)(x) + 2(x)(−2y) + 2(2)(−2y)

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

= 4 + x2 + 4y2 + 4x − 4xy − 8y

(2 + x − 2y)2 = 4 + x2 + 4y2 + 4x − 4xy − 8y

(vi) We have,

(a2 + b2 + c2)2 = (a2)2 + (b2)2 + (c2)2 + 2a2b2 + 2b2c2 + 2a2c2

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

(a2 + b2 + c2)2 = a4 + b4 + c4 + 2a2b2 + 2b2c2 + 2c2a2

(vii) We have,

(ab + bc + ca)2 = (ab)2 + (bc)2 + (ca)2 + 2(ab)(bc) + 2(bc)(ca) + 2(ab)(ca)

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

= a2b2 + b2c2 + c2a2 + 2(ac)b2 + 2(ab)(c)2 + 2(bc)(a)2

(ab + bc + ca)2 = a2b2 + b2c2 + c2a2 + 2acb2 + 2abc2 + 2bca2

(viii) We have,

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

(ix) We have,

(a/bc + b/ca + c/ab)2 = (a/bc)2 + (b/ca)2 + (c/ab)2 + 2(a/bc)(b/ca) + 2(b/ca)(c/ab) + 2(a/bc)(c/ab)

[∴ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

(x) We have,

(x + 2y + 4z)2 = x2 + (2y)2 + (4z)2 + 2x × 2y + 2 × 2y × 4z + 2x × 4z

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

(x + 2y + 4z)2 = x2 + 4y2 + 16z2 + 4xy + 16yz + 8xz

(xi) We have,

(2x − y + z)2 = (2x)2 + (−y)2 + (z)2 + 2(2x)(−y) + 2(− y)(z) + 2(2x)(z)

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

(2x − y + z)2 = 4x2 + y2 + z2 − 4xy − 2yz + 4xz

(xii) We have,

(−2x + 3y + 2z)2 = (−2x)2 + (3y)2 + (2z)2 + 2(−2x)(3y) + 2(3y)(2z) + 2(−2x)(2z)

[∴ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

(−4x + 6y + 4z)2 = 4x2 + 9y2 + 4z2 − 12xy + 12yz − 8xz

Question: 2

Use algebraic identities to expand the following algebraic equations.

(i)  (a + b + c)2 + (a − b + c)2

(ii) (a + b + c)2 − (a − b + c)2

(iii) (a + b + c)2 + (a + b − c)2 + (a + b − c)2

(iv) (2x + p − c)2 − (2x − p + c)2

(v) (x2 + y2 + (−z)2) − (x2 − y2 + z2)2

Solution:

(i) We have,

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

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

= 2a2 + 2b2 + 2c2 + 4ca

(a + b + c)2 + (a − b + c)2 = 2a2 + 2b2 + 2c2 + 4ca

(ii)  We have,

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

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

= a2 + b2 + c2 + 2ab + 2bc + 2ca − a2 − b2 − c2 + 2ab + 2bc − 2ca)

= 4ab + 4bc

(a + b + c)2 − (a − b + c)2 = 4ab + 4bc

(iii) We have,

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

= a2 + b2 + c2 + 2ab + 2bc + 2ca + (a2 + b2 + (z)2 − 2bc − 2ab + 2ca) + (a2 + b2 + c2 − 2ca − 2bc + 2ab)

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

= 3a2 + 3b2 + 3c2 + 2ab + 2bc + 2ca − 2bc − 2ab − 2ca − 2bc + 2ab

= 3x2 + 3y2 + 3z2 + 2ab − 2bc + 2ca

(a + b + c)2 + (a + b − c)2 + (a − b + c)2 = 3a2 + 3b2 + 3c2 + 2ab − 2bc + 2ca

(a + b + c)2 + (a + b − c)2 + (a − b + c)2 = 3(a2 + b2 + c2) + 2(ab − bc + ca)

(iv) We have,

(2x + p − c)2 − (2x − p + c)2

= [2x2 + p2 + (−c)2 + 2(2x)p + 2p(−c) + 2(2x)(−c)] − [4x2 + (−p)2 + c2 + 2(2x)(−p) + 2c(−p) + 2(2x)c]

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

(2x + p − c)2 − (2x − p + c)2 = [4x2 + p2 + c2 + 4xp − 2pc − 4xc] − [4x2 + p2 + c2 − 4xp − 2pc + 4xc]

Opening the bracket,

(2x + p − c)2 − (2x − p + c)2 = 4x2 + p2 + c2 + 4xp − 2pc − 4cx − 4x2 − p2 − c2 + 4xp + 2pc − 4cx]

(2x + p − c)2 − (2x − p + c)2 = 8xp − 8xc

= 8x(p − c)

Hence, (2x + p − c)2 − (2x − p + c)2 = 8x(p − c)

(v) We have,

(x2 + y2 + (−z)2)2 − (x2(−y)2 + z2)2

= [x4 + y4 + (-z)4 + 2x2y2 + 2y2(−z)2 + 2x2(- z)2] − [x4 + (−y)4 + z4 − 2x2y2 − 2y2z2 + 2x2z2]

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

Taking the negative sign inside,

= [x4 + y4 + (−z)4 + 2x2y2 + 2y2(−z)2 + 2x2(−z)2] − [x4 + (−y)4 + z4 − 2x2y2 − 2y2z2 + 2x2z2]

= 4x2y- 4z2x2

Hence, (x2 + y2 + (− z)2)2−(x2(−y)2 + z2)2 = 4x2y2 - 4z2x2

Question: 3

If a + b + c = 0 and a2 + b2 + c2 = 16, find the value of ab + bc + ca:

Solution:

We know that,

[∵ (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]

(0)2 = 16 + 2(ab + bc + ca)

2(ab + bc + ca)= -16

ab + bc + ca =-8

Hence, value of required express ab + bc + ca = - 8

Question: 4

If a2 + b2 + c2 = 16 and ab + bc + ca = 10, find the value of a + b + c?

Solution:

We know that,

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

(x + y + z)2 = 16 + 2(10)

(x + y + z)2 = 36

(x + y + z) = √36

(x + y + z) = ± 6

Hence, value of required expression I; (a + b + c) = ± 8

Question: 5

If a + b + c = 9 and ab + bc + ca = 23, find value of a2 + b2 + c2

Solution:

We know that,

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

92 = a2 + b2 + c2 + 2(23)

81 = a2 + b2 + c2 + 46

a2 + b2 + c2 = 81 − 46

a2 + b2 + c2 = 35

Hence, value of required expression a2 + b2 + c2 = 35

Question: 6

Find the value of the equation: 4x2 + y2 + 25z2 + 4xy − 10yz − 20zx when x = 4, y = 3, z = 2

Solution:

4x2 + y2 + 25z2 + 4xy − 10yz − 20zx

(2x)2 + y2 + (−5z)2 + 2(2x)(y) + 2(y)(−5z) + 2(−5z)(2x)

(2x + y − 5z)2

(2(4) + 3 − 5(2))2

(8 + 3 − 10)2

(1)2

1

Hence value of the equation is equals to 1

Question: 7

Simplify each of the following expressions:

(i) (x + y + z)2 + (x + y/2 + 2/3)2 − (x/2 + y/3 + z/4)2

(ii) (x + y − 2z)2 − x2 − y2 − 3z2 + 4xy

(iii) [x2 − x + 1]2 − [x2 + x + 1]2

Solution:

(i) Expanding, we get

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

Rearranging coefficients,

(ii) Expanding, we get

(x + y − 2z)2 − x2 − y2 − 3z2 + 4xy

= [x2 + y2 + 4z2 + 2xy + 2y(−2z) + 2a(−2c)] − x2 − y2 − 3z2 + 4xy

= z2 + 6xy − 4yz − 4zx

(x + y − 2z)2 − x2 − y2 − 3z2 + 4xy = z2 + 6xy − 4yz − 4zx

(iii) Expanding, we get

[x2 − x + 1]2 − [x2 + x + 1]2

= (x2)2 + (- x)2 + 12 + 2(x2)(−x) + 2(−x)(1) + 2x2) − [(x2)2 + x2 + 1 + 2x2x + 2x(1) + 2x2(1)]

[∵ (x + y + z)2 = x2 + y2 + z2 + 2xy + 2yz + 2xz]

= x4 + y2 + 1 − 2x3 − 2x + 2x2 − x2 − x4 − 1 − 2x3 − 2x − 2x2

= −4x3 − 4x

= −4x(x2 + 1)

Hence simplified equation = [x2 − x + 1]2 − [x2 + x + 1]2 = −4x(x2 + 1)

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