{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 128 128 128 1 0 0 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times " 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 265 49 "High School Modul es > Algebra by Gregory A. Moore" }}{PARA 3 "" 0 "" {TEXT -1 4 " " }{TEXT 264 21 "Completing the Square" }}{PARA 0 "" 0 "" {TEXT -1 120 " \nThe square root principle, the method of completing the square and i ts application to solving second degree equations.\n" }}{PARA 0 "" 0 " " {TEXT 266 153 "[Directions : Execute the Code Resource section first . Although there will be no output immediately, these definitions are \+ used later in this worksheet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {SECT 0 {PARA 4 "" 0 "" {TEXT -1 7 "0. Code" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 605 "SquarerootSolve := proc(eq)\n local EQ,EQ1, EQ2,L,R,s1,s2;\n \+ print(`\\nOriginal Problem :`); print(eq);\n\n EQ := sqrt(lhs(eq )) = sqrt(rhs(eq)); \n print(`\\nTake squareroot of each side`); pr int(EQ);print(` `);\n print(`\\nDon't forget the +/-`); \n print (lhs(EQ), ` = (+/-) `, rhs(EQ));print(` `);\n\n L := op(1,op(1,simp lify( abs(sqrt(lhs(EQ))))));\n R := rhs(EQ); \n\n EQ1 := L = R; \+ EQ2 := L = -R;\n print(`\\nNow break this into two linear equation s`);\n print( EQ1,EQ2);\n\n s1 := solve(EQ1,x); s2 := solve(EQ2, x);\n print(`\\nThe solutions are : `); print( s1, s2 );\nend proc :\n\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1130 "CompleteSteps := proc(EQ) #ONLY for leading coefficient = +1 or -1\n local EQ2 ,EQ3,C,b, lt ;\n\n print(`\\nOriginal Equation :`); print(EQ);\n\n E Q2 := sort(op(1,EQ)-op(2,EQ), x);\n \n C := coeff(EQ2,x,0);\n EQ3 : = EQ2 - C = - C;\n print(`\\n1. Put x terms on left and the number te rm on right :`);\n print(EQ3); print(` `);\n\n lt := coeff(EQ2,x,2); lt:= lt/abs(lt);\n if( lt = -1) then\n EQ3 := lt*lhs(EQ3) = lt* rhs(EQ3);\n print(`(If the coefficient of x^2 is negative then mu ltiply both sides by -1)`);\n print(EQ3); \n fi;\n\n b := coeff ( lhs(EQ3), x);\n print(`\\n2. Take half of the x coefficinet`,b,\n \+ ` and square it, add both sides`);\n print(EQ3); \n print(cat( `+ `, convert((b/2)^2, string),\n ` +`, convert((b/2) ^2, string) )); print(`_______ ___`); \n EQ2 := lhs(EQ3) + (b/2)^2 = rhs(EQ3) + (b/2)^2;\n print(EQ2);\n\n print(` `); \n print(factor( lhs(EQ2)) = rhs(EQ2));\n\n factor(lhs(EQ2)) = rhs(EQ2);\n sqrt(lhs(% )) = sqrt(rhs(%));\n print(`\\n3. Take the square root of each side ( with + or -)`);\n print(`\\n4. Now solve the two linear equations :`) ; print(solve(%, x));\nend proc:\n\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 1977 "CompleteStepsMore := proc(EQ) #ONLY for case when \+ leading coef\037\255 1\n local EQ2,EQ3,A,B,C,b, lt, sg, subEQ, rEQ,g \+ ;\n \n\n print(`\\nOriginal Equation :`); print(EQ);\n subEQ := sor t(op(1,EQ)-op(2,EQ), x);\n A := coeff(subEQ,x,2); B := coeff(subEQ,x, 1); C := coeff(subEQ,x,0);\n g := A;\n if( B <> 0 ) then g := gcd( g , B); fi;\n if( C <> 0 ) then g := gcd( g, C); fi;\n \n EQ3 := (sub EQ - C)/g = - C/g; \n \n lt := coeff((subEQ/g),x,2);\n if(lt=1) th en CompleteSteps(EQ3);\n else\n \n print(` `);\n print(`\\n1 . Factor out any common terms and put x terms on left and the number t erm on right :`);\n print(EQ3); print(` `);\n A := A/g; sg := A /abs(A);\n if( sg = -1) then \n EQ3 := -lhs(EQ3) = -rhs(EQ3); \n print(`\\n(If the coefficient of x^2 is negative, then multipl y both sides by -1)`);\n print(EQ3); \n fi;\n\n lt := abs(A ); \n subEQ := lhs(EQ3)/lt; rEQ := rhs(EQ3);\n print(`\\n Fact or the leading coefficient from the left side`);\n print(cat(lt,`(`,s ubEQ,`) = `, rEQ));\n \n\n b := coeff( subEQ, x);\n print(`\\n2. Ta ke half of the x coefficient`,b,\n ` and square it to get `, ( b/2)^2);\n print(`..then MULTIPLY it by leading coefficient factor`,l t,\n ` and add to both sides :`);\n \n subEQ := subEQ + (b/2) ^2; \n print(cat(\n lt,`(`,subEQ,`) = `,rEQ,` + `,lt,`*`, (b/2)^2, `)`\n ));\n rEQ := rEQ + lt*(b/2)^2; \n\n EQ2 := \+ lt*subEQ = rEQ;\n\n print(`\\n3. Take the square root of each side (w ith +/-)`);\n EQ2 := factor(lhs(EQ2)) = rhs(EQ2); print(EQ2);\n\n \n EQ3 := sqrt(lhs(%)) = sqrt(rhs(%));\n lt:= op(1, lhs(EQ3)); \+ rEQ := rhs(EQ3);\n subEQ := op(1, op(2, lhs(EQ2)));\n \n print(` ` ); print(`\\n4. Now solve the two linear equations :`); \n print(cat( lt,`(`, subEQ ,`) = `, rEQ )); \n print(cat( lt,`(`, \+ convert(subEQ, string) ,`) = `, -rEQ )); \n print(` `);\n print ( lt*subEQ = rEQ);\n print( lt*subEQ = -rEQ);\n print(` `);\n\n \n \+ print(x = \{solve(EQ2, x)\});\n fi;\nend proc:" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 28 "1. The Square Root Principle" }}{PARA 0 "" 0 "" {TEXT -1 129 "Here is a straight forward problem. You can probably gue ss the answers; but, lets use this as an example of this solution meth od." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SquarerootSolve(x^2 = 9);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "SquarerootSolve(x^2 = 324);" }}}{PARA 0 "" 0 "" {TEXT -1 98 "\n\nThat works fine enough f or perfect squares. How about for numbers which are NOT perfect square s?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "SquarerootSolve(x^2 = \+ 8);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "SquarerootSolve(x^2 \+ = 75);" }}}{PARA 0 "" 0 "" {TEXT -1 52 "\nWhat if the expression being squared is not just x?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "S quarerootSolve( (x-3)^2 = 25 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "SquarerootSolve( (3*x+2)^2 = 40 );" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 36 "SquarerootSolve( ((x+1)/2)^2 = 49 );" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 55 "2. The Idea of Completing the Square (lea ding coef = 1)" }}{PARA 0 "" 0 "" {TEXT -1 39 "\nFirst note the differ ence between the " }{TEXT 256 16 "complete squares" }{TEXT -1 37 " of \+ binomials and incomplete squares." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "for k from 1 to 10 do \n `Perfect Square = `, (x +k)^2=expand((x+k)^2), \n ` Incomplete Square = `, expand( \+ (x+k)^2 - k^2); od;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "for k from 1 to 10 do \n `Perfect Square = `, (x-k)^2=expand((x-k)^2) , \n ` Incomplete Square = `, expand( (x-k)^2 - k^2); od;" }}}{PARA 0 "" 0 "" {TEXT -1 200 "\nThe incomplete squares are \"missin g\" the constant term. We can make these \"incomplete squares\" become complete by adding the term that will make them complete. But how do \+ we figure out what that is?\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "(x + a)^2: % = expand(%);\n" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{PARA 0 "" 0 "" {TEXT -1 43 "Notice that the last term is the square of " }{TEXT 260 1 "a" }{TEXT -1 41 " , while the middle coefficient i s twice " }{TEXT 261 1 "a" }{TEXT -1 1 "!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "(x + 7)^2: % = expand (%);" }}}{PARA 0 "" 0 "" {TEXT -1 127 "\nNotice that the last term is \+ the square of 7 , while the middle coefficient is twice 7. Therefore, \+ if we new the middle term, " }{TEXT 259 2 "2a" }{TEXT -1 33 ", we coul d divide it by 2 to get " }{TEXT 258 1 "a" }{TEXT -1 240 ", and then s quare a to get the \"missing\" constant term. \n\n____________________ _____________________________________________________________\nLets tr y an experiment with this idea. We'll start with an incomplete square \+ and try to complete it." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "E xpression := x^2 + 18*x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "`The middle coefficient is `; b := coeff(Expression, x, 1);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "a := b/2; a2 := a^2;" }}} {PARA 0 "" 0 "" {TEXT -1 112 "\n\nNow if we add the square of a to the original expression and try to factor it, we SHOULD get a perfect squ are!" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Expression + a2: % = factor(%);" }}}{PARA 0 "" 0 "" {TEXT -1 50 "And we do! It works!\n\nL ets try some more examples." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "x^2 + 24*x;\na := coeff(%, x, 1)/2;\n%% + %^2 = factor( %% + %^2); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 65 "x^2 + 100*x;\na := coef f(%, x, 1)/2;\n%% + %^2 = factor( %% + %^2);" }}}{PARA 0 "" 0 "" {TEXT -1 53 "\nIt also works if the middle coefficient is negative." } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "x^2 - 30*x;\na := coeff(%, \+ x, 1)/2;\n%% + %^2 = factor( %% + %^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "x^2 - 84*x;\na := coeff(%, x, 1)/2;\n%% + %^2 = facto r( %% + %^2);" }}}{PARA 0 "" 0 "" {TEXT -1 82 "\nIt even works if the \+ middle term is odd! However, a will end up being a fraction." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "x^2 + 5*x;\na := coeff(%, x, 1)/2;\n%% + %^2 = factor( %% + %^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "x^2 + 21*x;\na := coeff(%, x, 1)/2;\n%% + %^2 = facto r( %% + %^2);" }}}{PARA 0 "" 0 "" {TEXT -1 122 "\nSo thats it. We halv e then middle coefficient, then square it and add that number to make \+ an incomplete square, complete." }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 64 "3. Solving Equations by Completing the Square (leading coef = 1)" }}{PARA 0 "" 0 "" {TEXT -1 3 "To " }{TEXT 257 19 "complete the square " }{TEXT -1 138 " on an equation which is not in \"square = square\" f orm, we need to do a little preparation, and then use the method above . First, we write" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "Complet eSteps( x^2 + 6*x - 7= 0);" }}}{PARA 0 "" 0 "" {TEXT -1 228 "\nYou may notice that the last problem is actually much more easily solved by s imply factoring the original equation. However, this method also works where factoring fails, and gives us answers not possible via integer \+ factoring." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "CompleteSteps( x^2 + 6*x + 7= 0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "Comp leteSteps( x^2 + 20*x + 36 = 0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "CompleteSteps( x^2 - 20*x + 36 = 0);" }}}{PARA 0 "" 0 "" {TEXT -1 79 "\nIt also works when there answers are not real. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "CompleteSteps( x^2 + 8*x - 11= 0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "CompleteSteps( x^2 + 6*x + 20= 0);" }}}{PARA 0 "" 0 " " {TEXT -1 167 "\n\nNote that the problem might be slightly obscured t o begin with because the terms are not in the \"right places.\" We can easily fix that by putting them back in order." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 30 "CompleteSteps( x + 10 = x^2);" }}}{PARA 0 "" 0 "" {TEXT -1 96 "\n\nSometimes we have an additional step - to make s ure that the coefficient on x^2 is positive 1." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "CompleteSteps( 2 - x^2 = x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "CompleteSteps( 8 - x^2 = 5*x);" }}}{PARA 0 "" 0 "" {TEXT -1 62 "\n\nThere is also a built-in function for complet ing the square." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "student[c ompletesquare](x^2 + 6*x = 7); \nsolve(%, x);" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 55 "4. More Difficult Equations to solve (leading coef \+ \255 1)" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "CompleteStepsMore ( 2*x^2 + 3*x + 1 = 0); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "CompleteStepsMore( 3 - 2*x = 4*x^2); solve( 3 - 2*x = 4*x^2);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "CompleteStepsMore( 5*x^2 + x - 1 = 0); solve(5*x^2 + x - 1 = 0, x);" }}}{PARA 0 "" 0 "" {TEXT -1 83 "\nSometimes, the equation only appears to have a leading coefficie nt other than one." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "Comple teStepsMore( 100*x^2 + 200*x + 100 = 0);" }}}{PARA 0 "" 0 "" {TEXT -1 23 "\nLets try this big guy." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "CompleteStepsMore( 3600*x^2-15120*x+15876=0 );" }}}{PARA 0 "" 0 " " {TEXT -1 23 "\nNot so big after all!\n" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 37 "5. Equations of Parabolas and Circles" }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }{TEXT 263 9 "PARABOLAS" }{TEXT -1 370 "\n\nIts very n ice to express parabolas in the form : y = (x-a)^2 + b or (y-b) = (x-a )^2, because we can immediately read the vertex as being (a,b). This m akes it easy to graph and easy to find the maximum or minimum point of the parabola. When we are given an equation for a parabola which is N OT in the form, our new friend \"completing the square\" can come to t he rescue." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "y = x^2 + 6*x \+ + 7;\nstudent[completesquare](%,x); \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "y = x^2 + 46*x ;\nstudent[completesquare](%,x);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "y = x^2 + 9*x + 3;\nstudent[ completesquare](%,x); " }}}{PARA 0 "" 0 "" {TEXT -1 40 "\nThis works f or downward parabolas also." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "y = -x^2 + 20*x - 23;\nstudent[completesquare](%,x); " }}}{PARA 0 "" 0 "" {TEXT -1 182 "\nThis also works for parabolas which have some \+ vertical stretching or compression. These more generalized parabolas a re expressable in the form y = c(x-a)^2 + b, or (y-b) = c(x-a)^2." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "y = 4*x^2 + 9*x - 11;\nstude nt[completesquare](%,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "y = 1205*x^2 + 4030*x ;\nstudent[completesquare](%,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "y = -20000*x^2 + 926*x -8;\nstudent [completesquare](%,x);" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n\n" }{TEXT 262 7 "CIRCLES" }{TEXT -1 320 "\n\nTo find the equation of a circle in center-radius form - (x-h)^2 + (y-k)^2 = r^2 - its sometimes necessa ry to perform completion of the square on BOTH x and y! Although it m ay appear to be double trouble, its only the same process performed tw ice.\n\nHere is an example. First we complete the square with respect \+ to x." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "x^2 + 6*x + y^2 -8* y = 7;\nstudent[completesquare](%,x);" }}}{PARA 0 "" 0 "" {TEXT -1 173 "\nTo tidy things up a bit, lets add 9 to both sides. This gets al l of the loose constants onto the right side, where they are safely aw ay from dangerously spinning variables." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "lhs(%) + 9 = rhs(%) + 9;" }}}{PARA 0 "" 0 "" {TEXT -1 56 "\nNow lets do it again - but this time with respect to y." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "student[completesquare](%,y) ;" }}}{PARA 0 "" 0 "" {TEXT -1 19 "\nTidy up a bit ...." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "lhs(%) + 16 = rhs(%) + 16; " }}} {PARA 0 "" 0 "" {TEXT -1 90 "\n\nThere it is! The center is (-3,4) and the radius is sqrt(32).\n\nLets try another example." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "x^2 - 200*x + y^2 + 200*y = 0;\nstudent[c ompletesquare](%,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "lhs (%) + 10000 = rhs(%) + 10000;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "student[completesquare](%,y);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "lhs(%) + 10000 = rhs(%) + 10000; " }}}}{PARA 0 "" 0 " " {TEXT 267 36 "\n \251 2002 Waterloo Maple Inc " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{MARK "0 1" 49 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }