<Text-field style="Heading 2" layout="Heading 2">0. C\303\263digo</Text-field>

14-Completando o Quadrado

Estudaremos o princ\303\255pio da raiz quadrada, o m\303\251todo de completar o quadrado e a sua aplica\303\247\303\243o na resolu\303\247\303\243o de equa\303\247\303\265es de segundo grau.

restart;

Solu\303\247\303\243oraizquadrada := proc(eq) local EQ,EQ1, EQ2,L,R,s1,s2; print(`\134nProblema Original:`); print(eq); EQ := sqrt(lhs(eq)) = sqrt(rhs(eq)); print(`\134nPegue o quadrado de cada superf\303\255cie `); print(EQ);print(` `); print(`\134nN\303\243o esque\303\247a o +/-`); print(lhs(EQ), ` = (+/-) `, rhs(EQ));print(` `); L := op(1,op(1,simplify( abs(sqrt(lhs(EQ)))))); R := rhs(EQ); EQ1 := L = R; EQ2 := L = -R; print(`\134nAgora +quebre isto em duas equa\303\247\303\265es lineares`); print( EQ1,EQ2); s1 := solve(EQ1,x); s2 := solve(EQ2,x); print(`\134nAs solu\303\247\303\265es s\303\243o: `); print( s1, s2 ); end proc:

Completandoquadrado := proc(EQ) #ONLY for leading coefficient = +1 or -1 local EQ2,EQ3,C,b, lt ; print(`\134nEqua\303\247\303\243o Original:`); print(EQ); EQ2 := sort(op(1,EQ)-op(2,EQ), x); C := coeff(EQ2,x,0); EQ3 := EQ2 - C = - C; print(`\134n1. Coloque os termos x a esquerda e os termos num\303\251ricos a direita:`); print(EQ3); print(` `); lt := coeff(EQ2,x,2); lt:= lt/abs(lt); if( lt = -1) then EQ3 := lt*lhs(EQ3) = lt*rhs(EQ3); print(`(se o coeficiente de x^2 \303\251 negativo multiplique os dois lados por -1)`); print(EQ3); fi; b := coeff( lhs(EQ3), x); print(`\134n2. Pegue a metade do coeficiente x `,b, ` e eleve ao quadrado, adicione para ambos os lados`); print(EQ3); print(cat(`+ `, convert((b/2)^2, string), ` +`, convert((b/2)^2, string) )); print(`_______ ___`); EQ2 := lhs(EQ3) + (b/2)^2 = rhs(EQ3) + (b/2)^2; print(EQ2); print(` `); print(factor(lhs(EQ2)) = rhs(EQ2)); factor(lhs(EQ2)) = rhs(EQ2); sqrt(lhs(%)) = sqrt(rhs(%)); print(`\134n3. Tome o quadrado de cada lado (com + ou -)`); print(`\134n4. Agora resolva as duas equa\303\247\303\265es lineares:`); print(solve(%, x)); end proc:

Completandoquadradocomplicado := proc(EQ) #ONLY for case when leading coef\037\302\255 1 local EQ2,EQ3,A,B,C,b, lt, sg, subEQ, rEQ,g ; print(`\134nEqua\303\247\303\243o Original:`); print(EQ); subEQ := sort(op(1,EQ)-op(2,EQ), x); A := coeff(subEQ,x,2); B := coeff(subEQ,x,1); C := coeff(subEQ,x,0); g := A; if( B <> 0 ) then g := gcd( g, B); fi; if( C <> 0 ) then g := gcd( g, C); fi; EQ3 := (subEQ - C)/g = - C/g; lt := coeff((subEQ/g),x,2); if(lt=1) then CompleteSteps(EQ3); else print(` `); print(`\134n1. Fatore cada termo comum e coloque os termos x a esquerda e os termos num\303\251ricos a direita:`); print(EQ3); print(` `); A := A/g; sg := A/abs(A); if( sg = -1) then EQ3 := -lhs(EQ3) = -rhs(EQ3); print(`\134n(Se o coeficiente de x^2 \303\251 negativo, ent\303\243o multiplique os dois lados por -1)`); print(EQ3); fi; lt := abs(A); subEQ := lhs(EQ3)/lt; rEQ := rhs(EQ3); print(`\134n Fatore o coeficiente principal a partir do lado esquerdo`); print(cat(lt,`(`,subEQ,`) = `, rEQ)); b := coeff( subEQ, x); print(`\134n2. Pegue a metade do coeficiente x `,b, ` e eleve ao quadrado para obter `, (b/2)^2); print(`..Agora multiplique isto pelo fator do coeficiente principal `,lt, ` e adicione isso para os dois lados:`); subEQ := subEQ + (b/2)^2; print(cat( lt,`(`,subEQ,`) = `,rEQ,` + `,lt,`*`,(b/2)^2, `)` )); rEQ := rEQ + lt*(b/2)^2; EQ2 := lt*subEQ = rEQ; print(`\134n3. Pegue o quadrado de cada lado (com +/-)`); EQ2 := factor(lhs(EQ2)) = rhs(EQ2); print(EQ2); EQ3 := sqrt(lhs(%)) = sqrt(rhs(%)); lt:= op(1, lhs(EQ3)); rEQ := rhs(EQ3); subEQ := op(1, op(2, lhs(EQ2))); print(` `); print(`\134n4. Agora resolva as duas equa\303\247\303\265es lineares:`); print(cat( lt,`(`, subEQ ,`) = `, rEQ )); print(cat( lt,`(`, convert(subEQ, string) ,`) = `, -rEQ )); print(` `); print( lt*subEQ = rEQ); print( lt*subEQ = -rEQ); print(` `); print(x = {solve(EQ2, x)}); fi; end proc:

<Text-field style="Heading 2" layout="Heading 2">1. Princ\303\255pio da Raiz Quadrada</Text-field>

Esse \303\251 um problema bem direto. Primeiro voc\303\252 deve ser capaz de adivinhar as respostas; mas vamos estimular aqui um m\303\251todo de resolu\303\247\303\243o.

SquarerootSolve(x^2 = 9);

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SquarerootSolve(x^2 = 324);

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Isso funciona bem com quadrados perfeitos. E quanto aos n\303\272meros que n\303\243o s\303\243o quadrados perfeitos?

SquarerootSolve(x^2 = 8);

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SquarerootSolve(x^2 = 75);

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e se a express\303\243o elevada ao quadrado n\303\243o for apenas x?

SquarerootSolve( (x-3)^2 = 25 );

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SquarerootSolve( (3*x+2)^2 = 40 );

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SquarerootSolve( ((x+1)/2)^2 = 49 );

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<Text-field style="Heading 2" layout="Heading 2">2. Completando o Quadrado (coeficiente principal = 1)</Text-field>

Inicialmente, observemos a diferen\303\247a entre quadrados completos de bin\303\264mios e quadrados incompletos. Nos quadrados incompletos falta um termo constante.

Exemplo1 :

for k from 1 to 10 do `Quadrado Perfeito = `, (x+k)^2=expand((x+k)^2), ` Quadrado Incompleto = `, expand( (x+k)^2 - k^2); od;

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for k from 1 to 10 do `Quadrado Perfeito = `, (x-k)^2=expand((x-k)^2), ` Quadrado Incompleto = `, expand( (x-k)^2 - k^2); od;

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A quest\303\243o \303\251 como fazer como que esses quadrados incompletos se tornem completos; como descobrir o termo que os fa\303\247a completos?

(x + a)^2: % = expand(%);

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Veja que o \303\272ltimo termo \303\251 o quadrado de a, enquanto o ceficiente do meio \303\251 duas vezes o a!

(x + 7)^2: % = expand(%);

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Perceba que o \303\272ltimo termo \303\251 o quadrado de 7 , enquanto o coeficiente do meio \303\251 duas vezes 7. Portanto, se soubermos o novo termo do meio, 2a, podemos dividi-lo por 2 para achar a, e elev\303\241-lo ao quadrado para obter a constante que precisamos. _________________________________________________________________________________ Agora, vamos fazer uma experi\303\252ncia com essa id\303\251ia. Vamos come\303\247ar com um quadrado incompleto e ent\303\243o tentar complet\303\241-lo.

Express\303\243o := x^2 + 18*x;

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`O coeficiente m\303\251dio \303\251 `; b := coeff(Expression, x, 1);

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a := b/2; a2 := a^2;

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Agora, se somarmos o quadrado de a \303\240 express\303\243o original e tentarmos fator\303\241-la, devemos obter um quadrado perfeito!

Expression + a2: % = factor(%);

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E encontramos! Funciona! Vejamos outros exemplos.

x^2 + 24*x; a := coeff(%, x, 1)/2; %% + %^2 = factor( %% + %^2);

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x^2 + 100*x; a := coeff(%, x, 1)/2; %% + %^2 = factor( %% + %^2);

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Tamb\303\251m funciona se o coeficiente do meio for negativo.

x^2 - 30*x; a := coeff(%, x, 1)/2; %% + %^2 = factor( %% + %^2);

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x^2 - 84*x; a := coeff(%, x, 1)/2; %% + %^2 = factor( %% + %^2);

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Funciona at\303\251 se o coeficiente do meio for \303\255mpar! No entanto, a fatalmente ser\303\241 uma fra\303\247\303\243o

x^2 + 5*x; a := coeff(%, x, 1)/2; %% + %^2 = factor( %% + %^2);

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x^2 + 21*x; a := coeff(%, x, 1)/2; %% + %^2 = factor( %% + %^2);

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Ent\303\243o \303\251 isso. Temos agora o coeficiente do meio; ent\303\243o, elevamos ao quadrado e somamos esse n\303\272mero para transformar um quadrado incompleto em completo.

<Text-field style="Heading 2" layout="Heading 2">3. Resolvendo Equa\303\247\303\265es Completando o Quadrado (coeficiente principal)</Text-field>

Para completar o quadrado em uma equa\303\247\303\243o que n\303\243o est\303\241 na forma "quadrado = quadrado", temos de fazer uma pequena prepara\303\247\303\243o e ent\303\243o usar o m\303\251todo acima. Primeiro, escrevemos:

CompleteSteps( x^2 + 6*x - 7= 0);

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Voc\303\252 pode notar que o \303\272ltimo problema \303\251 mais facilmente solucion\303\241vel se fatorarmos a equa\303\247\303\243o original. No entanto, esse m\303\251todo tamb\303\251m funciona onde a fatora\303\247\303\243o falha e nos d\303\241 respostas que n\303\243o s\303\243o poss\303\255veis atrav\303\251s da fatora\303\247\303\243o de n\303\272meros inteiros.

CompleteSteps( x^2 + 6*x + 7= 0);

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CompleteSteps( x^2 + 20*x + 36 = 0);

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CompleteSteps( x^2 - 20*x + 36 = 0);

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Ele tamb\303\251m funciona quando as respostas n\303\243o s\303\243o reais. <Intermediate Algebra use only>

CompleteSteps( x^2 + 8*x - 11= 0);

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CompleteSteps( x^2 + 6*x + 20= 0);

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Observe que o problema pode ser um pouco complicado de interpretar porque os termos n\303\243o est\303\243o nos seus "devidos lugares." Podemos consertar isso facilmente, colocando-os em ordem.

CompleteSteps( x + 10 = x^2);

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As vezes, temos um passo adicional - para verificar se o coeficiente de x^2 \303\251 1 positivo.

CompleteSteps( 2 - x^2 = x);

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CompleteSteps( 8 - x^2 = 5*x);

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Temos tamb\303\251m uma fun\303\247\303\243o embutida para completar o quadrado.

student[completesquare](x^2 + 6*x = 7); solve(%, x);

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<Text-field style="Heading 2" layout="Heading 2">4. Equa\303\247\303\265es Mais Dif\303\255ceis de Resolver (coeficiente principal)</Text-field>

CompleteStepsMore( 2*x^2 + 3*x + 1 = 0);

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CompleteStepsMore( 3 - 2*x = 4*x^2); solve( 3 - 2*x = 4*x^2);

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CompleteStepsMore( 5*x^2 + x - 1 = 0); solve(5*x^2 + x - 1 = 0, x);

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Certas vezes, a equa\303\247\303\243o parece ter o coeficiente principal diferente de um.

CompleteStepsMore( 100*x^2 + 200*x + 100 = 0);

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Vamos agora tentar essa aqui :

CompleteStepsMore( 3600*x^2-15120*x+15876=0 );

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<Text-field style="Heading 2" layout="Heading 2">5. Equa\303\247\303\265es de Par\303\241bolas e C\303\255rculos</Text-field>

PAR\303\201BOLAS Para limitar o trabalho \303\251 bom expressarmos as par\303\241bolas na forma: y = (x-a)^2 + b ou (y-b) = (x-a)^2, porque podemos imediatamente ler o v\303\251rtice sendo (a,b). Isso facilita a cria\303\247\303\243o do gr\303\241fico e determinar o ponto m\303\241ximo e o ponto m\303\255nimo da par\303\241bola. Quando temos uma equa\303\247\303\243o de uma par\303\241bola que n\303\243o est\303\241 nessa forma, o nosso novo amigo "completando o quadrado" vem nos salvar.

Exemplos :

y = x^2 + 6*x + 7; student[completesquare](%,x);

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y = x^2 + 46*x ; student[completesquare](%,x);

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y = x^2 + 9*x + 3; student[completesquare](%,x);

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Tamb\303\251m funciona com par\303\241bolas negativas :

y = -x^2 + 20*x - 23; student[completesquare](%,x);

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Isso tamb\303\251m funciona com par\303\241bolas que tenham esticamento ou compress\303\243o vertical. Essas par\303\241bolas menos espec\303\255ficas s\303\243o expressas na forma y = c(x-a)^2 + b, ou (y-b) = c(x-a)^2.

y = 4*x^2 + 9*x - 11; student[completesquare](%,x);

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y = 1205*x^2 + 4030*x ; student[completesquare](%,x);

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y = -20000*x^2 + 926*x -8; student[completesquare](%,x);

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C\303\215RCULOS Para encontramos a equa\303\247\303\243o de um c\303\255rculo de raio na forma - (x-h)^2 + (y-k)^2 = r^2 - certas vezes \303\251 necess\303\241rio completar os quadrados tanto de x quanto de y! Apesar de parecer ser complicado, n\303\243o passa do simples processo que acabamos de efetuar. Veja esse exemplo. Primeiro completamos o quadrado em rela\303\247\303\243o a x.

x^2 + 6*x + y^2 -8*y = 7; student[completesquare](%,x);

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Para arrumar um pouco as coisas, vamos somar 9 aos dois lados. Vamos colocar todas as constantes no lado direito, onde elas ficar\303\243o a salvo de certas vari\303\241veis intrometidas.

lhs(%) + 9 = rhs(%) + 9;

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Agora vamos repetir o processo - Mas desta vez em rela\303\247\303\243o a y :

student[completesquare](%,y);

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Arrumando as coisas ....

lhs(%) + 16 = rhs(%) + 16;

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E pronto! O centro \303\251 (-3,4) e o raio \303\251 a reaiz quadrada de (32). Vamos tentar outro exemplo.

x^2 - 200*x + y^2 + 200*y = 0; student[completesquare](%,x);

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lhs(%) + 10000 = rhs(%) + 10000;

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student[completesquare](%,y);

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lhs(%) + 10000 = rhs(%) + 10000;

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