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]
= 4x2y2 - 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)