Question

The method of completing the square is used several times in this text. It is a useful method for combining several quadratic and linear polynomials into a perfect square plus a constant. Prove the following identity, which is one general form of completing the square:

\(\begin{array}{c}\sum\limits_{i = 1}^n {{a_i}{{\left( {x - {b_i}} \right)}^2} + cx} \\\,\,\,\, = \left( {\sum\limits_{i = 1}^n {{a_i}} } \right){\left( {x - \frac{{\sum\limits_{i = 1}^n {{a_i}{b_i} - \frac{c}{2}} }}{{\sum\limits_{i = 1}^n {{a_i}} }}} \right)^2}\\\,\,\,\,\,\, + {\sum\limits_{i = 1}^n {{a_i}\left( {{b_i} - \frac{{\sum\limits_{i = 1}^n {{a_i}{b_i}} }}{{\sum\limits_{i = 1}^n {{a_i}} }}} \right)} ^2}\\\,\,\,\,\, + {\left( {\sum\limits_{i = 1}^n {{a_i}} } \right)^{ - 1}}\left[ {c\sum\limits_{i = 1}^n {{a_i}{b_i} - \frac{{{c^2}}}{4}} } \right]\\if\,\,\sum\limits_{i = 1}^n {{a_i} \ne 0.} \end{array}\)

Short answer

Proved.

Step-by-step solution

Given information

Completing the square is a useful method for combining several quadratic and linear polynomials into a perfect square plus constant.

Proving part

First, expand the left-hand side of the equation to get,

\(\begin{aligned}\sum\limits_{i = 1}^n {{a_i}{{\left( {x - {b_i}} \right)}^2} + cx} &= cx + \sum\limits_{i = 1}^n {{a_i}\left[ {{x^2} - 2{b_i}x + {b_i}^2} \right]} \\ &= cx + \sum\limits_{i = 1}^n {\left[ {{a_i}{x^2} - 2{a_i}{b_i}x + {a_i}{b_i}^2} \right]} ...\left( 1 \right)\end{aligned}\)

Now collect all the square terms in x.

The coefficient of \({x^2}\) is \(\sum\limits_{i = 1}^n {{a_i}} \)

The coefficient of x is \(c - 2\sum\limits_{i = 1}^n {{a_i}{b_i}} \)

The constant term is \(\sum\limits_{i = 1}^n {{a_i}b_i^2} \)

So,

\({x^2}\sum\limits_{i = 1}^n {{a_i} + x\left[ {c - 2\sum\limits_{i = 1}^n {{a_i}{b_i}} } \right] + \sum\limits_{i = 1}^n {{a_i}b_i^2} } ...\left( 2 \right)\)

Next, expand each term on the right side of the original equation to produce

Then,

\(\begin{aligned}\sum\limits_{i = 1}^n {{a_i}{{\left( {x - {b_i}} \right)}^2} + cx} &= \left( {\sum\limits_{i = 1}^n {{a_i}} } \right)\left( {{x^2} - 2x{b_i} + b_i^2} \right) + cx\\ &= {x^2}\sum\limits_{i = 1}^n {{a_i} + x\left[ {c - 2\sum\limits_{i = 1}^n {{a_i}{b_i}} } \right] + \sum\limits_{i = 1}^n {{a_i}b_i^2} } \\ &= \left( {\sum\limits_{i = 1}^n {{a_i}} } \right)\left\{ {{x^2} - 2x\left( {\frac{{\sum\limits_{i = 1}^n {{a_i}{b_i} - \frac{c}{2}} }}{{\sum\limits_{i = 1}^n {{a_i}} }}} \right)} \right\}\\ &+ \sum\limits_{i = 1}^n {{a_i}b_i^2} + \frac{{{{\left( {\sum\limits_{i = 1}^n {{a_i}{b_i}} } \right)}^2}}}{{\sum\limits_{i = 1}^n {{a_i}} }} - \frac{{{{\left( {\sum\limits_{i = 1}^n {{a_i}{b_i}} } \right)}^2}}}{{\sum\limits_{i = 1}^n {{a_i}} }}\\ &- \frac{{c\sum\limits_{i = 1}^n {{a_i}{b_i}} }}{{\sum\limits_{i = 1}^n {{a_i}} }} + \frac{{\frac{{{c^2}}}{4}}}{{\sum\limits_{i = 1}^n {{a_i}} }} + \frac{{c\sum\limits_{i = 1}^n {{a_i}{b_i}} }}{{\sum\limits_{i = 1}^n {{a_i}} }} - \frac{{\frac{{{c^2}}}{4}}}{{\sum\limits_{i = 1}^n {{a_i}} }}\\ &= \left( {\sum\limits_{i = 1}^n {{a_i}} } \right)\left( {{x^2} - 2x\frac{{\sum\limits_{i = 1}^n {{a_i}{b_i} - \frac{c}{2}} }}{{\sum\limits_{i = 1}^n {{a_i}} }}} \right) + \frac{{{{\left( {\sum\limits_{i = 1}^n {{a_i}{b_i}} } \right)}^2} - c\sum\limits_{i = 1}^n {{a_i}{b_i}} + \frac{{{c^2}}}{4}}}{{\sum\limits_{i = 1}^n {{a_i}} }}\\\,\,\,\,\,\, &+ \sum\limits_{i = 1}^n {{a_i}b_i^2 - \frac{{{{\left( {\sum\limits_{i = 1}^n {{a_i}{b_i}} } \right)}^2}}}{{\sum\limits_{i = 1}^n {{a_i}} }} + \frac{{c\sum\limits_{i = 1}^n {{a_i}{b_i} - \frac{{{c^2}}}{4}} }}{{\sum\limits_{i = 1}^n {{a_i}} }}} \end{aligned}\)

Hence, [Proved].