SECTION A.3 Polynomials A29 6 Complete the Square The idea behind completing the square in one variable is to “adjust” an expression of the form + x bx 2 to make it a perfect square. Perfect squares are trinomials of the form ( ) ( ) + + = + − + = − x ax a x a x ax a x a 2 or 2 2 2 2 2 2 2 For example, + + x x6 9 2 is a perfect square because ( ) + + = + x x x 6 9 3 . 2 2 And − + p p 12 36 2 is a perfect square because ( ) − + = − p p p 12 36 6 . 2 2 So how do we “adjust” + x bx 2 to make it a perfect square? We do it by adding a number. For example, to make + x x6 2 a perfect square, add 9. But how do we know to add 9? If we divide the coefficient of the first-degree term, 6, by 2, and then square the result, we obtain 9. This approach works in general. Completing the Square of + x bx 2 • Identify the coefficient of the first-degree term, namely b . • Multiply b by 1 2 and then square the result. That is, compute ( )b 1 2 . 2 • Add ( )b 1 2 2 to + x bx 2 to get ( ) ( ) + + = + x bx b x b 1 2 2 2 2 2 CAUTION To use ( )b1 2 2 to complete the square, the coefficient of the x2 term must be 1. j Figure 22 y 4 Area = y2 Area = 4y Area = 4y 4 y Completing the Square Determine the number that must be added to each expression to complete the square. Then factor the expression. Start Add Result Factored Form + y y8 2 ( ) ⋅ = 1 2 8 16 2 + + y y8 16 2 ( ) + y 4 2 + x x 12 2 ( ) ⋅ = 1 2 12 36 2 + + x x 12 36 2 ( ) + x 6 2 20 2 − a a ( ) ( ) ⋅ − = 1 2 20 100 2 20 100 2 − + a a 10 2 ( ) −a − p p5 2 ( ) ( ) ⋅ − = 1 2 5 25 4 2 − + p p5 25 4 2 ( ) − p 5 2 2 EXAMPLE 10 Notice that the factored form of a perfect square is either ( ) ( ) ( ) ( ) + + = + − + = − x bx b x b x bx b x b 2 2 or 2 2 2 2 2 2 2 2 Now Work problem 125 Are you wondering why we refer to making an expression a perfect square as “completing the square”? Look at the square in Figure 22. Its area is ( ) +y 4 . 2 The yellow area is y2 and each brown area is y4 (for a total area of y8 ). The sum of these areas is + y y8 2 . To complete the square, we need to add the area of the green region, which is ⋅ = 4 4 16. As a result, ( ) + + = + y y y 8 16 4 . 2 2
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