9.2 Solving Quadratic Equations by Completing the Square

1 2001 McGraw-Hill Companies679 Solving Quadratic Equations by Completing the a Quadratic equation by the squareroot a Quadratic equation by completingthe a geometric application involving aquadratic equationIn Section , we solved Quadratic Equations by factoring and using the zero productrule. However, not all Equations are factorable over the integers. In this section, we willlook at another method that can be used to solve a Quadratic equation, called the squareroot , we will solve a special type of equation using the factoring methodof Chapter Equations by FactoringSolve the Quadratic equation x2 16by write the equation in standard form:x2 16 0 Factoring, we have(x 4)(x 4) 0 Finally, the solutions arex 4orx 4or 4 Example 1 NOTEHere, we factor thequadratic member of theequation as a difference YOURSELF 1 Solve each of the following Quadratic Equations .(a)5x2 180(b)x2 25 The equation in Example 1 could have been solved in an alternative fashion.

2 We could haveused what is called the Square root , given the equationx2 16we can write the equivalent statementx orx This yields the solutionsx 4orx 4or 4 This discussion leads us to the following general sure to include boththe positive and the negativesquare roots when you use thesquare root , FUNCTIONS, ANDINEQUALITIES 2001 McGraw-Hill CompaniesExample 2 further illustrates the use of this the Square Root MethodSolve each equation by using the Square root method.(a)x2 9By the Square root property,x orx 3 3or 3 (b)x2 17 0 Add 17 to both sides of the 17so x or or , (c)2x2 3 02x2 3x2 x x or(d)x2 1 0x2 1x x ior i 1 1 162 162A32321171171171171919 Example 2 NOTEIf a calculator wereused, (rounded tothree decimal places).117 Example 2(d ) we seethat complex-number solutionsmay x2 k, when kis a complex number, thenx orx 2k2kRules and Properties: Square Root PropertyWe can also use the approach in Example 2 to solve an equation of the form(x 3)2 16 CHECK YOURSELF 2 Solve each equation.

3 (a)x2 5(b)x2 2 0(c)3x2 8 0(d)x2 9 0 SOLVINGQUADRATICEQUATIONS BYCOMPLETING 2001 McGraw-Hill CompaniesUsing the Square Root MethodUse the Square root method to solve each equation.(a)(x 5)2 5 0(x 5)2 5x 5 x 5 or 5 (b)3(y 1)2 2 03(y 1)2 2(y 1)2 y 1 y 1 orThe approximate solutions are , . 3 163 3 163163A2323151515 Example 3 NOTEThe two solutionsand areabbreviated as 5 . Using acalculator, we find theapproximate solutions , .155 155 15 NOTEWe have solved for yand rationalized we combine the terms onthe right, using the commondenominator of 1213 12 1313 13 163 CHECK YOURSELF 3 Using the Square root method, solve each equation.(a)(x 2)2 3 0(b)2(x 1)2 1As before, by the Square root property we havex 3 4 Subtract 3 from both sides of the for xyieldsx 3 4which means that there are two solutions:x 3 4orx 3 4 1 7or 1, 7 Not all Quadratic Equations can be solved directly by factoring or using the Square rootmethod. We must extend our Square root method is useful in this process because any Quadratic equation can bewritten in the form(x h)2 kwhich yields the solutionx h 1kNOTEIf (x h)2 k, thenx h andx h 1k1k682 CHAPTER9 QUADRATICEQUATIONS, FUNCTIONS, ANDINEQUALITIES 2001 McGraw-Hill CompaniesThe process of changing an equation in standard formax2 bx c 0to the form(x h)2 kis called the method of Completing the Square ,and it is based on the relationship betweenthe middle term and the last term of any perfect- Square s look at three perfect- Square trinomials to see whether we can detect a pattern:x2 4x 4 (x 2)2(1)x2 6x 9 (x 3)2(2)x2 8x 16 (x 4)2(3)Note that in each case the last (or constant) term is the Square of one-half of the coefficientof xin the middle (or linear) term.

4 For example, in equation (2),x2 6x 9 (x 3)2of this coefficient is 3, and ( 3)2 9, the this relationship for yourself in equation (3). To summarize, in perfect- Square trino-mials, the constant is always the Square of one-half the coefficient of are now ready to use the above observation in the solution of Quadratic Equations bycompleting the Square . Consider Example Completing the Square to Solve an EquationSolve x2 8x 7 0 by Completing the , we rewrite the equation with the constant on the right-hand side:x2 8x 7 Our objective is to have a perfect- Square trinomial on the left-hand side. We know that wemust add the Square of one-half of the xcoefficient to complete the Square . In this case, thatvalue is 16, so now we add 16 to each side of the 8x 16 7 16 Factor the perfect- Square trinomial on the left, and combine like terms on the right to yield(x 4)2 23 Now the Square root property yieldsx 4 Subtracting 4 from both sides of the equation givesx 4 or 4 As decimals, these solutions are approximated by.

5 123123123 NOTEN otice that thisrelationship is true onlyif theleading, or x2, coefficient is will be important 4 NOTE12 8 4 and 42 16 NOTEWhen you graph therelated function,y x2 8x 7, you will notethat the xvalues for the xintercepts are just below 1 andjust above 9. Be certain thatyou see how these points relateto the exact solutions, 4 and 4 .123123 CHECK YOURSELF 4 Solve x2 6x 2 0by Completing the that if(x h)2 k, then x h .1kSOLVINGQUADRATICEQUATIONS BYCOMPLETING 2001 McGraw-Hill CompaniesCompleting the Square to Solve an EquationSolve x2 5x 3 0 by Completing the 5x 3 0 Add 3 to both 5x 3 Make the left-hand side a perfect 5x 3 Take the Square root of both Solve for orThe approximate solutions are , . 5 1372 5 1372137252374 x 52 2 52 2 52 2 Example 5 Completing the Square to Solve an EquationSolve x2 4x 13 0 by Completing the 4x 13 0 Subtract 13 from both 4x 13 Add to both 4x 4 13 4 Factor the left-hand side.(x 2)2 9 Take the Square root of both 2 Simplify the 2 x 2 3ix 2 3ior 2 3i 19i1 9B12 (4)R2 Example 6 CHECK YOURSELF 5 Solve x2 3x 7 0by Completing the YOURSELF 6 Solve x2 10x 41 Equations have nonreal complex solutions, as Example 6 the Square of one-half of the xcoefficient to bothsides of the equation.

6 Note that12 5 52 NOTEN otice that the graph ofy x2 4x 13 does notintercept the 7 illustrates a situation in which the leading coefficient of the Quadratic mem-ber is not equal to 1. As you will see, an extra step is , FUNCTIONS, ANDINEQUALITIES 2001 McGraw-Hill CompaniesCompleting the Square to Solve an EquationSolve 3x2 6x 7 0 by Completing the 6x 7 0 Add 7 to both 6x 7 Divide both sides by 2x Now, complete the Square on the 2x 1 1 The left side is now a perfect Square .(x 1)2 x 1 x 1 3 1303A103A1031037373 Example 7 CHECK YOURSELF 7 Solve 2x2 8x 3 0 by Completing the following algorithm summarizes our work in this section with Solving quadraticequations by Completing the 1 Isolate the constant on the right side of the 2 Divide both sides of the equation by the coefficient of the x2term ifthat coefficient is not equal to 3 Add the Square of one-half of the coefficient of the linear term to bothsides of the equation. This will give a perfect- Square trinomial on theleft side of the 4 Write the left side of the equation as the Square of a binomial, andsimplify on the right 5 Use the Square root property, and then solve the resulting by Step: Completing the SquareCAUTIONB efore you can complete thesquare on the left, thecoefficient of x2must be equalto 1.

7 Otherwise, we must divideboth sides of the equation bythat have rationalizedthe denominator and combinedthe terms on the right s proceed now to applications involving the Completing of a SquareThe length of a rectangle is 4 cm greater than its width. If the area of the rectangle is 108 cm2,what are the approximate dimensions of the rectangle?Step 1 You are asked to find the dimensions (the length and the width) of the 8 SOLVINGQUADRATICEQUATIONS BYCOMPLETING 2001 McGraw-Hill CompaniesCHECK YOURSELF 8In a triangle, the base is 4in. less than its height. If its area is , find thelength of the base and the height of the the Completing of a SquareAn open box is formed from a rectangular piece of cardboard, whose length is 2 in. morethan its width, by cutting 2-in. squares from each corner and folding up the sides. If the vol-ume of the box is to be 100 , what must be the size of the original piece of cardboard?Step 1We are asked for the dimensions of the sheet of 2 Again sketch the 222222222xExample 9 Step 2 Whenever geometric figures are involved in an application, start by drawing,and then labeling,a sketch of the problem.

8 Letting xrepresent the width and x 4 thelength, we haveStep 3 The area of a rectangle is the product of its length and width, sox(x 4) 108 Step 4x(x 4) 108x2 4x 108x2 4x 4 108 4(x 2)2 112x 2 x 2 Step 5We reject (cm) as a solution. A length cannot be negative, and sowe must consider only (cm) in finding the required width xis approximately cm, and the length x 4 is cm. Because ( cm)( cm) gives a rectangle of area cm2, the solution is verified. 2 1112 2 111211121112xWidthx 4 LengthNOTEM ultiply and completethe , FUNCTIONS, ANDINEQUALITIES 2001 McGraw-Hill CompaniesStep 3To form an equation for volume, we sketch the completed volume is the product of height, length, and width,2(x 2)(x 4) 100 Step 42(x 2)(x 4) 100(x 2)(x 4) 50x2 6x 8 50x2 6x 42x2 6x 9 42 9(x 3)2 51x 3 x 3 Step 5 Again, we need consider only the positive solution. The width xof the originalpiece of cardboard is approximately in., and its length x 2 is in. Thedimensions of the completed box will be by by 2 in.

9 , which gives volume of anapproximate 100 both sides by 2, and multiply on the left. Then solveas 2 (Length)x 4 (Width)2(Height)NOTEThe original width ofthe cardboard was x. Removingtwo 2-in. squares leaves x 4for the width of the , the length of the boxis x 2. Do you see why?CHECK YOURSELF 9A similar box is to be made by cutting 3-cm squares from a piece of cardboard that is4cm longer than it is wide. If the required volume is 300cm3, find the dimensionsof the original sheet of YOURSELF ANSWERS1. (a) 6, 6 ; (b) 5, 5 2. (a) , ; (b) , ; (c); (d) 3i, 3i 3. (a) 2 ; (b)4. 3 5 4i in.; height cm cm 4 1102 3 1372 111 2 122 13 2163, 2163 12121515 2001 McGraw-Hill CompaniesExercisesIn exercises 1 to 8, solve by factoring or Completing the 6x 5 5x 6 2z 35 5q 24 5x 3 10x 8 y 2 3z 1 0In exercises 9 to 20, use the Square root method to find solutions for the 12 66 12 5 3217.(x 1)2 1218.(2x 3)2 519.(2z 1)2 3 020.(3p 4)2 9 0 Name Section Date 2001 McGraw-Hill CompaniesIn exercises 21 to 32, find the constant that must be added to each binomial expression toform a perfect- Square 4ayIn exercises 33 to 54, solve each equation by Completing the 12x 2 14x 7 2y 4z 72 2x 5 2x 10x 13 3x 17 5z 7 8q 20 m 3 y 5 x x 2x 1 3x 6x 8x 1 2x 12 2y 3 8x 20 2x 10 0In exercises 55 to 60, find the constant that must be added to each binomial to form aperfect- Square trinomial.

10 Let xbe the variable; other letters represent 4abxIn exercises 61 and 62, solve each equation by Completing the 2ax 2ax 8 0 Solve the following width of a rectangle is 3 ft less than its length. If the area of the rectangle is70 ft2, what are the dimensions of the rectangle? length of a rectangle is 5 cm more than its width. If the area of the rectangle is84 cm2, find the dimensions of the length of a rectangle is 2 cm more than 3 times its width. If the area of therectangle is 85 cm2, find the dimensions of the the length of a rectangle is 3 ft less than twice its width and the area of therectangle is 54 ft2, what are the dimensions of the rectangle? length of a rectangle is 1 cm more than its width. If the length of the rectangle isdoubled, the area of the rectangle is increased by 30 cm2. What were the dimensionsof the original rectangle? 2001 McGraw-Hill 2001 McGraw-Hill box is to be made from a rectangular piece of tin that is twice as long as it is accomplish this, a 10-cm Square is cut from each corner, and the sides are foldedup.