Express A^2 + B^2 + C^2 – Ab – Bc – Ca As Sum Of Squares

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Express a\(^{2}\) + b\(^{2}\) + c\(^{2}\) - ab - bc - ca as Sum of Squares

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)\(^{2}\)]

9th Grade Math

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