Completing the Square

The starting position for each of the above cases was thecorrect choice of the complete square. For example, part (a)began with the expansion of ( x + 3) as x + 6 x + 9 and this information was used tocomplete the square on x+6 x+10. In practice the starting point isusually the quadratic expression, i.e., in (a) it would be x+ 6 x + 10. The problem is to work directlyfrom this. The general procedure uses the following observation:

so the expression x + px can be made into a complete square byadding ( p/2), i.e., by adding the square of half thecoeffcient of x. So as not to alter the expression the sameamount must be subtracted. In other words we use the equality

and this is the essence of completing the square.

Example 1

Complete the square on the following expressions.

(a) x + 8 x + 15 , (b) x - 5 x + 6 .

Solution

(a) Completing the square means adding the square of half thecoefficient of x and then subtracting the same amount. Thus

x + 8 x + 15 = [ x + 8 x] + 15 ,

= [( x + 4) - 4 ] + 15 ,

= ( x + 4) - 16 + 15 ,

= ( x + 4) - 1 .

(b) For x - 5x + 6 the procedure is much the same.

x - 5x + 6 = [x - 5x] + 6 ,