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 Depdendent Variable

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 Dependent Variable

 Number of inequalities to solve: 23456789
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## Completing the Square

The starting position for each of the above cases was the correct 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 to complete the square on x+6 x+10. In practice the starting point is usually the quadratic expression, i.e., in (a) it would be x+ 6 x + 10. The problem is to work directly from this. The general procedure uses the following observation:

so the expression x + px can be made into a complete square by adding ( p/2), i.e., by adding the square of half the coeffcient of x. So as not to alter the expression the same amount 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 the coefficient 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 ,