{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 "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 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 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 281 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 286 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 287 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 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 288 49 "High School Modules > Algebra by Gregory A. Moore" }}{PARA 3 "" 0 "" {TEXT -1 4 " " }{TEXT 287 36 " The Quadratic Formula & Discriminant" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 64 "The derivation and use of the quadratic formula and discriminat." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT 289 153 "[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 1 " " }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 7 "0. Code" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 589 "Quadratic := proc( expr )\n local a, b, c, soln, k;\n a := coeff(expr,x,2); b := coeff(expr,x,1); c := coeff(expr,x,0);\n \+ printf(\"\\t\\t\\t\\t Quadratic Equation : \\t%A = 0\\n\",expr);\n \+ printf(\"\\t\\t\\t\\t Coefficients :\\t\\t\\t\\t\\t\\t\\t\\ta=%d\\t\\ t\\t b=%d\\t\\t\\t c=%d\\n\",\n a,b,c);\n soln := [ solve( expr = 0, x)];\n if(nops(soln) = 1) \n then printf(\"\\t \\t\\t\\t Solution : %A\\n\\n\",soln[1]);\n soln[1];\n \+ else printf(\"\\t\\t\\t\\t Solutions :\\t\\t\\t\\t\\t\\t\\t\\t\\t\\ t\\t%A\\t\\t%A\\n\\n\",\n soln[1],soln[2]);\n \+ soln[1],soln[2];\n fi;\n end proc:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 962 "Discriminant := proc( expr )\n local a, b, c, so ln, k, D;\n a := coeff(expr,x,2); b := coeff(expr,x,1); c := coeff( expr,x,0);\n printf(\"\\t\\t\\t\\t Quadratic Equation : \\t%A = 0\\ n\",expr);\n printf(\"\\t\\t\\t\\t Coefficients :\\t\\t\\t\\t\\t\\t \\t\\ta=%d\\t\\t\\t b=%d\\t\\t\\t c=%d\\n\",\n a,b,c) ;\n D := b*b - 4*a*c;\n printf(\"\\t\\t\\t\\t Discriminant =\\t\\ t\\t\\t\\t\\t\\t\\t%d\\n\",D);\n \n if( D > 0) \n then printf (\"\\t\\t\\t\\t There are two distinct real roots.\\n\\n\");\n \+ \n else if( D < 0) \n then printf(\"\\t\\t\\t\\t T here are two distinct complex roots.\\n\\n\");\n else printf (\"\\t\\t\\t\\t There is one real root. This quadratic expression is \+ a perfect square.\\n\\n\");\n fi;fi;\n if( D = floor(sqrt(D))^2) \+ \n then \n printf(\"\\t\\t\\t\\t This expression is factora ble. \\n\\n\"); print( factor(expr));\n else \n printf(\"\\t \\t\\t\\t This expression is NOT factorable. \\n\\n\");\n fi;\n p rint( solve( expr, x) );\n end proc:\n\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2083 "Compl eteStepsMore := proc(EQ) \n #ONLY for case when leading coe f\037\255 1 <>\n local EQ2,EQ3,A,B,C,b, lt, sg, subEQ, rEQ,g ;\n \n\n use RealDomain in\n\n print(`Origina l Equation :`); print(EQ);\n subEQ := sort(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 ) th en g := gcd( g, C); fi;\n \n EQ3 := (subEQ - C)/g = - C/g; \n \n \+ lt := coeff((subEQ/g),x,2);\n if(lt=1) then CompleteSteps(EQ3);\n el se\n \n print(`\\n1. Factor out any commong terms, and `,\n \+ `put x terms on left, number term 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(`(If the coefficient o f x^2 is negative,`,\n ` multiply both sides by -1)`);\n \+ print(EQ3); print(` `);\n fi;\n\n lt := abs(A); \n subEQ : = lhs(EQ3)/lt; rEQ := rhs(EQ3);\n print(`Factor the leading coeffi cient from the left side`);\n print(cat(lt,`(`,subEQ,`) = `, rEQ));\n \n b := coeff( subEQ, x);\n print(`\\n2. Take half of the x coeffici ent`,b,\n ` and square it to get `, (b/2)^2);\n print(`..then MULTIPLY it by leading coefficient factor`,lt,\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 prin t(` `); \n print(` `); print(`\\n3. Take the square root of each side (with +/-)`);\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 p rint( lt*subEQ = rEQ);\n print( lt*subEQ = -rEQ);\n print(` `);\n\n \+ \n print(x = \{solve(EQ2, x)\});\n fi;\nend use;\nend proc:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 " " {TEXT -1 28 "1. Derivation of the Formula" }}{PARA 0 "" 0 "" {TEXT -1 85 "\nYou might recall that we can solve any quadratic equation by \+ completing the square.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 " CompleteStepsMore( 3*x^2 + 30*x + 1 = 0);" }}}{PARA 0 "" 0 "" {TEXT -1 265 "\nEach of those problems is a bit long, and we seem to be doin g the same steps over and over each time. Is there a way to 'cut to th e chase' and get the result of that method faster? What if we did for \+ a general quadratic polynomial with letters instead of numbers.\n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "CompleteStepsMore( a*x^2 + b *x + c = 0);" }}}{PARA 0 "" 0 "" {TEXT -1 308 "\n\nWe get the formula \+ above. This is called the Quadratic Formula. It is usually written a l ittle differently in texts, as a single fraction with a 2a in the deno miator and a \"plus-or-minus\" in front of the squareroot to accomodat e both cases at once. It gives us a quick way of solving any quadratic formula." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "solve( a*x^2 + \+ b*x + c = 0, x);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 19 "2. Solving Problems" }}{PARA 0 "" 0 "" {TEXT -1 24 "\n Here is that formula.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "QuadForm := solve( a*x^2 + b*x + c = 0, x);" }}}{PARA 0 "" 0 "" {TEXT -1 21 "\n\nHere is an example." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "QuadEq := x^2 + 9*x + 20 = 0;" }}}{PARA 0 "" 0 "" {TEXT -1 116 "\nNow lets substitute for a, b, and c - the coefficients of x^2, x, and the constant term. Note that x^2 means 1*x^2." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "subs( \{ a = 1, b = 9, c = 2 0\}, \{QuadForm \});" }}}{PARA 0 "" 0 "" {TEXT -1 84 "\nWe can check o ur answer using Maple's solve command - and in this case, it factors. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "solve( QuadEq, x);\nfact or( lhs(QuadEq) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 141 "3*x^ 2 + 7*x + 4 ;\nsimplify( subs( \{ a = coeff(%,x,2) , b = coeff(%, x,1), \n c = coeff(%,x,0) \}, \{QuadForm \} ) );\n" }}}{PARA 0 "" 0 "" {TEXT -1 85 "\nHere is a little procesdure th at identifies the coefficients and finds the solutions" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "Quadratic( 5*x^2 + 7*x + 4);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "Quadratic( 6*x^2 - 216 );" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Quadratic( x^2 + 7*x );" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Quadratic( x^2 + 12 );" } }}{PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 19 "3. The Discriminant" }}{PARA 0 "" 0 "" {TEXT -1 20 "\nThe expre ssion D = " }{XPPEDIT 18 0 "b^2 - 4*a*c" "6#,&*$%\"bG\"\"#\"\"\"*(\"\" %F'%\"aGF'%\"cGF'!\"\"" }{TEXT -1 15 " is called the " }{TEXT 256 12 " discriminant" }{TEXT -1 113 " for a quadratic equation. You may recogn ize it as the quanitity under the radical sign in the quadratic formul a." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 " solve( a*x^2 + b*x + \+ c = 0, x);" }}}{PARA 0 "" 0 "" {TEXT -1 57 "\nWe can thus express the \+ quadratic formula in this form :" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Discr := b^2 -4*a*c; " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "x1 = (1/(2*a))*(-b + sqrt(Discr) );\nx2 = (1/(2*a))*(-b - sqrt (Discr) );" }}}{PARA 0 "" 0 "" {TEXT -1 124 "\n\nThe discriminant tell s us about the number and quality of roots of the equation. There are \+ three possibilities :\n " }{TEXT 265 1 "1" }{TEXT -1 7 ". D is " }{TEXT 266 8 "positive" }{TEXT -1 26 " the equation has " } {TEXT 267 3 "two" }{TEXT -1 1 " " }{TEXT 268 4 "real" }{TEXT -1 24 ", \+ but distint solutions\n" }{TEXT 269 10 " 2" }{TEXT -1 7 ". D i s " }{TEXT 270 4 "zero" }{TEXT -1 33 " the equation has \+ " }{TEXT 271 8 "one real" }{TEXT -1 19 " solution\n " }{TEXT 274 1 "3" }{TEXT -1 7 ". D is " }{TEXT 272 8 "negative" }{TEXT -1 26 " the equation has " }{TEXT 273 3 "two" }{TEXT -1 1 " " }{TEXT 275 7 "complex" }{TEXT -1 49 " solutions\n\nLets look at examples of \+ each case :" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Discriminant( x^2 + 3*x + 1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "Discrim inant( x^2 + 2*x + 1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "D iscriminant( x^2 + 1*x + 1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Discriminant( x^2 + 102*x + 1);" }}}{PARA 0 "" 0 "" {TEXT -1 328 "\nThe advantage to using the discriminant is that you can tell if there are real solutions or not simply by looking at the discriminant , instead of going through all of the work to solve the quadratic form ula.\n\nThe discriminant also tells us about the number of x intercept s for a graph. There are three possibilities :\n " }{TEXT 276 1 "1" }{TEXT -1 7 ". D is " }{TEXT 277 8 "positive" }{TEXT -1 26 " \+ the equation has " }{TEXT 278 3 "two" }{TEXT -1 1 " " }{TEXT 285 12 "x intercepts" }{TEXT -1 1 "\n" }{TEXT 279 10 " 2" }{TEXT -1 7 ". D is " }{TEXT 280 4 "zero" }{TEXT -1 33 " the eq uation has " }{TEXT 281 15 "one x intercept" }{TEXT -1 10 "\n \+ " }{TEXT 283 1 "3" }{TEXT -1 7 ". D is " }{TEXT 282 8 "negative" } {TEXT -1 26 " the equation has " }{TEXT 284 2 "no" }{TEXT -1 2 " " }{TEXT 286 12 "x intercepts" }{TEXT -1 58 "\n\nLets see graphs \+ of these three cases we looked at above." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "plot( x^2 + x - 5 , x = -6..6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "plot( 8*x^2 + 40*x + 50 , x = -6..3, y = -5.. 100);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "plot( x^2 + x + 2 \+ , x = -4..3, y = 0..10);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 17 "4. Factor ability " }}{PARA 0 "" 0 "" {TEXT -1 139 " \nThe discriminant also tel ls us another thing. A quadratic expression is factorable if, and only if the discriminant is a perfect square.\n" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 31 "Discriminant( (x+31)*(5*x-7) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "sqrt( 26244);" }}}{PARA 0 "" 0 "" {TEXT -1 55 "\nBut just adding one makes this expression unfactorable" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "Discriminant( (x+31)*(5*x-7) + 1 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "sqrt( 26224);" } }}{PARA 0 "" 0 "" {TEXT -1 124 "\nTry it for yourself. Any time you st art with a factorable quadratic expression, the discriminant will be a perfect square.\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "Discri minant( 8*x^2+62*x+117 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "sqrt( 100);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "Discrimi nant(102*x^2+145*x-408);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "sqrt(187489);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 " " 0 "" {TEXT -1 41 "5. Step by Step Method - Doing it by hand" }} {PARA 0 "" 0 "" {TEXT -1 71 "\nHere are steps you can use to solve the se problems by hand :\n " }{TEXT 257 1 "1" }{TEXT -1 28 ". Rea d a, b, and c\n " }{TEXT 258 1 "2" }{TEXT -1 33 ". Find the di scriminant\n " }{TEXT 259 1 "3" }{TEXT -1 47 ". Take the squar eroot of the discriminant\n " }{TEXT 260 5 " 4" }{TEXT -1 123 " . Form the two fractions\n\nThis method is slightly shorter and less p rone to errors than simply plugging into the formula.\n\n" }{TEXT 261 11 "Example 5.1" }{TEXT -1 3 ": " }{XPPEDIT 18 0 "5*x^2-8 = 0;" "6#/, &*&\"\"&\"\"\"*$%\"xG\"\"#F'F'\"\")!\"\"\"\"!" }{TEXT -1 2 " ." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 251 "#1 Read a, b, and c\n a: = 1; b:= 0; c:= -8;\n#2 Find the discriminate\n Discr := b^2 -4 *a*c;\n#3 Take the ssquareroot of the discriminant\n SqrtD := sqrt( Discr );\n#4 Form the fractions\n x = (1/(2*a))*(-b + SqrtD ), (1/ (2*a))*(-b - SqrtD );" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }{TEXT 263 11 "Example 5.2" }{TEXT -1 1 ":" }{TEXT 264 1 " " }{TEXT -1 1 " " } {XPPEDIT 18 0 "5*x^2 + 7*x - 2 = 0" "6#/,(*&\"\"&\"\"\"*$%\"xG\"\"#F' F'*&\"\"(F'F)F'F'F*!\"\"\"\"!" }{TEXT -1 2 " ." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 251 "#1 Read a, b, and c\n a:= 5; b:= 7; c:= - 2;\n#2 Find the discriminate\n Discr := b^2 -4*a*c;\n#3 Take the s squareroot of the discriminant\n SqrtD := sqrt( Discr );\n#4 Form t he fractions\n x = (1/(2*a))*(-b + SqrtD ), (1/(2*a))*(-b - SqrtD ) ;" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }{TEXT 262 11 "Example 5.3" } {TEXT -1 3 ": " }{XPPEDIT 18 0 "12*x^2+24*x+6 = 0;" "6#/,(*&\"#7\"\" \"*$%\"xG\"\"#F'F'*&\"#CF'F)F'F'\"\"'F'\"\"!" }{TEXT -1 2 " ." }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 252 "#1 Read a, b, and c\n a: = 12; b:= 24; c:= 6;\n#2 Find the discriminate\n Discr := b^2 - 4*a*c;\n#3 Take the ssquareroot of the discriminant\n SqrtD := sqrt ( Discr );\n#4 Form the fractions\n x = (1/(2*a))*(-b + SqrtD ), (1 /(2*a))*(-b - SqrtD );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT 290 35 "\n \251 2002 Waterloo Maple \+ Inc" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0" 30 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }