Here we will express a\(^{2}\) + b\(^{2}\) + c\(^{2}\) – ab – bc – ca as sum of squares.
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\(^{2}\) + b\(^{2}\) – 2ab) + (b\(^{2}\) + c\(^{2}\) – 2bc) + (c\(^{2}\) + a\(^{2}\) – 2ca)}
= \(\frac{1}{2}\){(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,
a\(^{2}\) + b\(^{2}\) + c\(^{2}\) – ab – bc – ca is always positive.
(ii) a\(^{2}\) + b\(^{2}\) + c\(^{2}\) – ab – bc – ca = 0 if \(\frac{1}{2}\){(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 4x\(^{2}\) + 9y\(^{2}\) + z\(^{2}\) – 6xy – 3yz – 2zx as sum of perfect squares.
Solution:
Given expression = 4x\(^{2}\) + 9y\(^{2}\) + z\(^{2}\) – 6xy – 3yz – 2zx
= (2x)\(^{2}\) + (3y)\(^{2}\) + z\(^{2}\) – (2x)(3y) – (3y)(z) – (z)(2x)
= ½[(2x - 3y)\(^{2}\) + (3y - z)\(^{2}\) + (z - 2x) \(^{2}\)].
2. If p\(^{2}\) + 4q\(^{2}\) + 25r\(^{2}\) = 2pq + 10qr + 5rp, prove that p = 2q = 5r.
Solution:
Here, p\(^{2}\) + 4q\(^{2}\) + 25r\(^{2}\) = 2pq + 10qr + 5rp
Or, p\(^{2}\) + 4q\(^{2}\) + 25r\(^{2}\) - 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 a\(^{2}\) + b\(^{2}\) + c\(^{2}\) - ab - bc - ca as Sum of Squares:
1. Express each of the following as a sum of perfect squares.
(i) x\(^{2}\) + y\(^{2}\) + z\(^{2}\) + xy + yz - zx
[Hint: Given expression = x\(^{2}\) + (-y)\(^{2}\) + z\(^{2}\) - x(-y) -(-y)z - zx
= ½[{x - (-y)}\(^{2}\) + {(-y) - z}\(^{2}\) + (z - x)\(^{2}\).]
(ii) 16a\(^{2}\) + b\(^{2}\) + 9c\(^{2}\) - 4ab - 3bc - 12ca
(iii) a\(^{2}\) + 25b\(^{2}\) + 4 - 5ab - 10b - 2a
2. If 4x\(^{2}\) + 9y\(^{2}\) + 16z\(^{2}\) - 6xy - 12yz - 8zx = 0, prove that 2x = 3y = 4z.
3. If a\(^{2}\) + b\(^{2}\) + 4c\(^{2}\) = 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)\(^{2}\)]
From Express a^2 + b^2 + c^2 – ab – bc – ca as Sum of Squares to HOME PAGE
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Nov 03, 24 12:50 PM
Oct 29, 24 01:27 PM
Oct 29, 24 12:21 AM
Oct 29, 24 12:06 AM
Oct 28, 24 12:53 AM
New! Comments
Have your say about what you just read! Leave me a comment in the box below. Ask a Question or Answer a Question.