Express a2 + b2 + c2 - ab - bc - ca as Sum of Squares
Here we will express a2 + b2 + c2 – ab – bc – ca as sum of squares.
a2 + b2 + c2 – ab – bc – ca = 12{2a2 + 2b2 + 2c2 – 2ab – 2bc – 2ca}
= 12{(a2 + b2 – 2ab) + (b2 + c2 – 2bc) + (c2 + a2 – 2ca)}
= 12{(a - b)2 + (b - c)2 + (c – a)2}
Corollaries:
(i) If a, b, c are real numbers then (a – b)2, (b – c)2 and (c – a)2 are positive as square of every real number is positive. So,
a2 + b2 + c2 – ab – bc – ca is always positive.
(ii) a2 + b2 + c2 – ab – bc – ca = 0 if 12{(a - b)2 + (b - c)2 + (c – a)2} = 0
Or, (a - b)2 = 0, (b - c)2 = 0, (c – a)2= 0
Or, a - b = 0, b - c = 0, c – a = 0, i.e., a = b = c
Solved Examples on Express a^2 + b^2 + c^2 – ab – bc – ca as Sum of Squares:
1. Express 4x2 + 9y2 + z2 – 6xy – 3yz – 2zx as sum of perfect squares.
Solution:
Given expression = 4x2 + 9y2 + z2 – 6xy – 3yz – 2zx
= (2x)2 + (3y)2 + z2 – (2x)(3y) – (3y)(z) – (z)(2x)
= ½[(2x - 3y)2 + (3y - z)2 + (z - 2x) 2].
2. If p2 + 4q2 + 25r2 = 2pq + 10qr + 5rp, prove that p = 2q = 5r.
Solution:
Here, p2 + 4q2 + 25r2 = 2pq + 10qr + 5rp
Or, p2 + 4q2 + 25r2 - 2pq - 10qr - 5rp = 0
Or, (p)2 + (2q)2 + (5r)2 – (p)(2q) – (2q)(5r) – (5r)(p) = 0
Or, ½[(p – 2q)2 + (2q – 5r)2 + (5r – p)2] = 0.
If sum of three positive numbers is zero, each number must be equal to 0.
Therefore, p – 2q = 0, 2q – 5r = 0, 5r – p = 0
Thus, p = 2q, 2q = 5r, 5r = p.
Therefore, p = 2q = 5r.
Practice Problems on Express a2 + b2 + c2 - ab - bc - ca as Sum of Squares:
1. Express each of the following as a sum of perfect squares.
(i) x2 + y2 + z2 + xy + yz - zx
[Hint: Given expression = x2 + (-y)2 + z2 - x(-y) -(-y)z - zx
= ½[{x - (-y)}2 + {(-y) - z}2 + (z - x)2.]
(ii) 16a2 + b2 + 9c2 - 4ab - 3bc - 12ca
(iii) a2 + 25b2 + 4 - 5ab - 10b - 2a
2. If 4x2 + 9y2 + 16z2 - 6xy - 12yz - 8zx = 0, prove that 2x = 3y = 4z.
3. If a2 + b2 + 4c2 = ab + 2bc + 2ca, prove that a = b = 2c.
Answers:
1. (i) ½[(x + y)2 + (y + z)2 + (z - x)2]
(ii) ½[(4a - b)2 + (b - 3c)2 + (3c - 4a)2]
(iii) ½[(a - 5b)2 + (5b - 2)2 + (2 - a)2]
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