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

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.



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.

Share this page: What’s this?

Recent Articles

  1. Types of Fractions |Proper Fraction |Improper Fraction |Mixed Fraction

    Jul 12, 24 03:08 PM

    Fractions
    The three types of fractions are : Proper fraction, Improper fraction, Mixed fraction, Proper fraction: Fractions whose numerators are less than the denominators are called proper fractions. (Numerato…

    Read More

  2. Worksheet on Fractions | Questions on Fractions | Representation | Ans

    Jul 12, 24 02:11 PM

    Worksheet on Fractions
    In worksheet on fractions, all grade students can practice the questions on fractions on a whole number and also on representation of a fraction. This exercise sheet on fractions can be practiced

    Read More

  3. Fraction in Lowest Terms |Reducing Fractions|Fraction in Simplest Form

    Jul 12, 24 03:21 AM

    Fraction 8/16
    There are two methods to reduce a given fraction to its simplest form, viz., H.C.F. Method and Prime Factorization Method. If numerator and denominator of a fraction have no common factor other than 1…

    Read More

  4. Conversion of Improper Fractions into Mixed Fractions |Solved Examples

    Jul 12, 24 12:59 AM

    To convert an improper fraction into a mixed number, divide the numerator of the given improper fraction by its denominator. The quotient will represent the whole number and the remainder so obtained…

    Read More

  5. Conversion of Mixed Fractions into Improper Fractions |Solved Examples

    Jul 12, 24 12:30 AM

    Conversion of Mixed Fractions into Improper Fractions
    To convert a mixed number into an improper fraction, we multiply the whole number by the denominator of the proper fraction and then to the product add the numerator of the fraction to get the numerat…

    Read More