SOLUCION DE ECUACIONES CUADRATICAS Y EL METODO DE CARDANO JOSE CUNDAPI Matricula: 2173046601
Septiembre 2018
´ Indice ´ 1 INTR INTRODUC ODUCCI CION
3
2 ECUACIONES CUADRATICAS
3
´ DE LA FORMULA GENERAL DE UN POLINOMIO 2.1 2.1 DEDUCCION DE GRADO 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 PROGRAM PROGRAMA A HECHO HECHO EN FOR FORTRAN TRAN QUE DETERMINA DETERMINA LAS RA´ICES DE UN POLINOMIO DE GRADO 2 . . . . . . . . . . . . . . . . . . . 2.2.1 ALGUNOS EJEMPLOS RESUELTOS . . . . . . . . . . . . . . . 2.3 PROGRAMA PARA COMPLEJOS . . . . . . . . . . . . . . . . . . . . .
3 METODO DE CARDANO
3 6 7 9 12
3.1 PROGRAMA CARDANO . . . . . . . . . . . . . . . . . . . . . . . . . . 4 BIBLIO BIBLIOGR GRAF AF´ IA
14 15
´ Indice de figuras 1
´ N NO LINEAL . . . . . . . . . . . . . . . . . . . . . . . . . . ECUACIO 1
7
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2 3 4 5
DOS DOS DOS DOS DOS DOS
RAICES REALES . . . . . . . . . . . . . . . . . . . . . . . . . RAICES IGUALES, DISCRIMINANTE=0 . . . . . . . . . . . . RAIC RAICES ES IMA IMAGINA GINARI RIAS AS,, DISC DISCRI RIMI MINA NANT NTE E MENO MENOR R QUE QUE 0 RAIC AICES IMA IMAGIN GINARIA ARIAS S US USA ANDO NDO EL PROG PROGR RAMA AMA 2 . . . .
. . . .
. . . .
8 8 9 11
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´ INT INTRODUCC UCCION
1
El planteamiento de un problema de la vida diaria requiere muchos casos para su soluci´ on, on, la representaci´on o n de n´ umeros umeros reales mediante s´ s´ımbolos hace posible encontrar los valores espec´ıficos ıficos de dichos s´ımbolos ımbolo s que satisfacen satisfa cen una relaci´on on de igualdad. De manera analoga podemos describir mediante ecuaciones el comportamiento de las sitaciones diarias como lo son el movimiento de un cuerpo conociendo un parametro temporal (el tiempo). Relaciones como lo son la velocidad, aceleraci´on,etc. on,etc. Ejemplo: [CONSIDERANDO EL M.R.U.A] 1 2
x = xi + vi + at2
(1)
´ RECORRIDA DEPENDE DEL TIEMPO, COMO ES UNA DONDE LA POSICION ´ DE GRADO PODRA TENER TRES POSIBLES RESULTADOS: ECUACION • Dos ra´ ra´ıces reales distintas. la par´abola abola corta el eje de las abscisas en dos puntos
diferentes. • Una ra´ ra´ız real, pero de multiplicidad dos o doble. La par´abola abola solo toca en un ´unico unico
punto al eje de las abscisas. • Dos Do s ra´ıces ıces complejas co mplejas conjugadas. conjuga das. La par´ pa r´abola abola no corta al eje de las abscisas.
´ EL TIPO DE RA´ICES DEPENDERA DEL DISCRIMINANTE DE LA ECUACI ON EL CUAL SE ABORDARA MAS ADELANTE.
2 2 .1
ECUA ECUACIONES CIONES CUADR CUADRA ATICAS TICAS ´ DEDUCCION DE LA FORMU ORMULA LA GENE GENERA RAL L DE UN POLINOMIO DE GRADO 2
Usando el m´etodo etodo de completaci´ on de cuadrados. Se demostrara que la soluci´on on o n de la ecuaci´ on on cuadr´atica atica es la famosa formula general. TENEMOS QUE LA ECUACION DE SEGUNDO GRADO ES DEL TIPO: ax2 + bx + c = 0
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CONSIDERANDO QUE a =0
PODEMOS DECIR QUE AL DESPEJAR c Y DIVIDIR ENTRE a, se sigue conservando la igualdad. igualdad. x2 +
Sumando
b c x = − a a 2
b 2a
En ambas lados de la igualdad se obtiene: b x2 + x + a
2
b 2a
c = − + a
2
b 2a
LO QUE ES EQUIVALENTE A:
b x + 2a
2
c = − + a
b2 4a2
REESCRIBIENDO EL TERMINO DE LA DERECHA: 2
+ = 2 x
b a
−4ac + b2 4a2
2
b x + 2a
=
b2 − 4ac 4a2
RESTANDO A TODA LA IGUALDAD EL LADO DERECHO:
b x + 2a
2
b2 − 4ac − =0 4a2
DANDOLE LA FORMA DE DIFERENCIA DE CUADRADOS:
b x + 2a
2
−
2
b2 − 4ac =0 4a2
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4 4 + 2 ) + + ) = 0 4 2 4 x
b a
−
b2 − ac a2
b2 − ac a2
b a
x
A PARTIR DE AQUI PARA QUE SE CUMPLA LA IGUALDAD PUEDE QUE:
4 + 2 = 0 4 x
´ A EXPRESION o´
−
b2 − ac a2
4 + 2 + 4 = 0 x
´ B EXPRESION
b a
b2 − ac a2
b a
RESOLVIENDO PARA LA EXPRESION A TENEMOS QUE:
( )=
( )=
b − − 2a
x A
b2 − 4ac 4a2
RESOLVIENDO PARA LA EXPRESION B TENEMOS QUE: x b
b − + 2a
b2 − 4ac 4a2
CON ESO OBTENEMOS LA FORMULA GENERAL COMO LA CONOCEMOS: x(A, B
4 )= 2 4 ( 4 )
x(A, B ) =
−
b ± a
−b ±
b2 − ac a2
b2 − ac
2a
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2.2
PROGR PROGRAM AMA A HEC HECHO HO EN EN FOR FORTR TRAN AN QUE QUE DET DETER ERMI MI-NA LAS RA´ ICES DE UN POLINOMIO DE GRADO 2
program program QUAD implicit implicit NONE real real * 4, parame parameter ter::z ::z = 1.e-6 1.e-6 double precision a,b,c,dis,x1,x2,a2 a,b,c,dis,x1,x2,a2 ´ write(*, write(*,*) *) ’PROGRAM ’PROGRAMA A QUE DETERMIN DETERMINA A LOS SOLUCION SOLUCIONES ES DE UNA ECUACI ECUACI ON ON DE SEGUND SEGUNDO O GRADO. GRADO.’ ’ !SOLICIT !SOLICITUD UD DE DATOS DATOS ´ PRINT PRINT *,’ESC *,’ESCRIB RIBE E LOS LOS VALORE VALORES S A,B,C A,B,C DE LA ECUAC ECUACIC IC ON ON DE SEGUND SEGUNDO O GRADO’ GRADO’ PRINT PRINT *,’DE *,’DE LA FORMA FORMA A*X^2+B* A*X^2+B*X+C= X+C=0’ 0’ READ READ (*,*) (*,*) a,b,c a,b,c PRINT PRINT *,’LOS *,’LOS VALORES VALORES LEIDOS LEIDOS SON:’ SON:’ WRITE (*,*)’A:’,a,’B:’,b (*,*)’A:’,a,’B:’,b,’C:’,c ,’C:’,c ! CALC CALCUL ULO O PARA PARA EL DISC DISCRI RIMI MINA NANT NTE E dis = b*b-4.0* b*b-4.0*a*c a*c a2 =2.0*a =2.0*a ! CONDICIONANDO CONDICIONANDO if (a==0)TH (a==0)THEN EN !CUANDO !CUANDO A=0 ´ write(*,*)’LA write(*,*)’LA ECUACI ON O N NO ES CUAD CUADRA RATI TICA CA SINO SINO LINE LINEAL AL’ ’ WRITE(*, WRITE(*,*)’L *)’LA A RAIZ ES:’,-c/ ES:’,-c/b b ELSEif(dis>=z)THEN ! se tendra tendra dos raice raices s reales reales distin distintas tas dis=sqrt(dis) x1=(-b+dis)/a2 x2=(-b-dis)/a2 write(*, write(*,*)’L *)’LA A ECUACION ECUACION TIENE TIENE DOS RAICES RAICES REALES REALES DISTINTA DISTINTAS:’ S:’ write(*, write(*,*)’x *)’x1=’, 1=’, x1, ’x2=’, ’x2=’, x2 ELSEIF ELSEIF (dis (dis <=-z) <=-z) THEN THEN ! EN ESTE ESTE CASO CASO HAY HAY DOS DOS RAIC RAICES ES COMP COMPLE LEJA JAS S dis=sqrt(-dis) x1=-b x1=-b /a2 x2=dis/a2 WRITE WRITE (*,*)’LA (*,*)’LA ECUACION ECUACION TIENE TIENE DOS RAICES RAICES COMPLEJA COMPLEJAS S CONJUGAD CONJUGADAS:’ AS:’ WRITE WRITE (*,*)’Pa (*,*)’Parte rte Real’, Real’, x1
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WRITE WRITE (*,* (*,*)’P )’Part arte e imagin imaginari aria’, a’,’ ’ ’,’+/’,’+/-’, ’, x2, x2, ’i’ ELSEIF(dis==0) ELSEIF(dis==0) THEN ! CASO EN EL QUE SOLO HAY UNA RAIZ x1=-b/a2 WRITE WRITE (*,*)’ (*,*)’LA LA ECUACI ECUACION ON TIENE TIENE UNA RAIZ RAIZ DOBLE’ DOBLE’ WRITE WRITE (*,*)’x= (*,*)’x=’, ’, x1 endif stop end progra program m QUAD QUAD 2.2.1 2.2.1
ALGUNOS ALGUNOS EJEMPLO EJEMPLOS S RESUE RESUEL LTOS
0x2 + x + 2 = 0
´ NO LINEAL Figura 1: ECUACION
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1x2 + 3x + 1 = 0
Figura 2: DOS RAICES REALES 1x2 + 2x + 1 = 0
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3x2 + 1x + 5 = 0
Figura 4: DOS RAICES IMAGINARIAS, DISCRIMINANTE MENOR QUE 0
2.3
PROG PROGRA RAMA MA PARA ARA COMP COMPLE LEJOS JOS
´ OTRO TAMBIEN SE ELABORO OTRO PROGRAM PROGRAMA A DONDE LAS ENTRADA ENTRADAS S SON NUMEROS COMPLEJOS, RECORDANDO QUE PARA EL INGRESO DE LOS MISMOS ES DE LA FORMA: Z = a + bi
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!COMPLEM !COMPLEMENTO ENTO PARA DETERMIN DETERMINAR AR LAS SOLUCION SOLUCIONES ES DE UNA ECUACION ECUACION ! CUADR CUADRATI ATICA CA CON ENTRA ENTRADAS DAS COMPL COMPLEJA EJAS S O REALES REALES. . program CmplxQuad CmplxQuad implicit implicit none complex::A,B,C,x,y,a1,a2 ´ WRITE(*, WRITE(*,*)’ *)’ ESTE PROGRAMA PROGRAMA DETERMIN DETERMINA A LAS SOLUCION SOLUCIONES ES DE UNA ECUACI ECUACI ON O N DE’ DE’ WRITE(*, WRITE(*,*)’S *)’SEGUN EGUNDO DO GRADO GRADO CON ENTRADAS ENTRADAS COMPLEJA COMPLEJAS S O REALES’ REALES’ WRITE(*,*)’’ write(*, write(*,*)’I *)’INGRE NGRESE SE LOS VALORES VALORES DE A,B,C’ A,B,C’ write( write(*,* *,*) ) ’El ’El ingres ingreso o de los datos datos debe ser de la forma’ forma’,’ ,’ ’,’z= ’,’z=a+b a+bi’ i’ WRIT WRITE( E(*, *,*) *)’( ’(VA VALO LOR R REAL REAL DE DE a, VAL VALOR OR DE DE b)’, b)’,’ ’ ’ WRITE(*, WRITE(*,*)’E *)’ENTRE NTRE PARENTES PARENTESIS IS SEPARADO SEPARADOS S POR COMA’ COMA’ write( write(*,* *,*)’P )’POR OR EJEMPL EJEMPLO’, O’,’ ’ ’,’(2. ’,’(2.0,0 0,0.0 .0)NU )NUMER MERO O 2’,’ o ’,’(2.0 ’,’(2.0,1. ,1.0) 0) Z=2+i’ Z=2+i’ write( write(*,* *,*) ) ’INGR ’INGRESE ESE LOS LOS DATOS DATOS ’ write(*, write(*,*)’ *)’ INGRESE INGRESE A’ READ*, READ*, A WRITE(*, WRITE(*,*) *) ’ASIGNAS ’ASIGNASTE TE A=’,A A=’,A write(*,*)’INGRESE write(*,*)’INGRESE B’ READ*, READ*, B WRITE(*, WRITE(*,*) *) ’ASIGNAS ’ASIGNASTE TE A=’,B A=’,B write(*,*)’INGRESE write(*,*)’INGRESE C’ READ*, READ*, C WRITE(*, WRITE(*,*) *) ’ASIGNAS ’ASIGNASTE TE C=’,C C=’,C a1=-b/(2*a) a2=sqrt(b*b-4.0*a*c) a2=0.5*a2/A x=a1+a2
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´ AHORA UTILIZANDO RESOLVIENDO OTRAVEZ LA SIGUIENTE ECUACION, EL PROGRAMA CON ENTRADAS COMPLEJAS O REALES. DE IGUAL MANERA CALCULA DE FORMA SATISFACTORIA LAS RAICES. 3x2 + 1x + 5 = 0
Figura 5: DOS RAICES IMAGINARIAS USANDO EL PROGRAMA 2
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3
MET METODO ODO DE CARD CARDAN ANO O
Cardano fue un ardiente astr´ologo, ologo, llevaba llevaba amuletos y predec´ predec´ıa el futuro durante las tormentas. Escribi´o muchos tratados sobre muchos temas, de medicina, religi´on, on, juegos, matem´ aticas. aticas. A sus 50 a˜ nos nos era er a un m´edico edico famoso fa moso y conocido. cono cido. Entre E ntre sus pacientes p acientes estuvo es tuvo el Arzobispo Arzobisp o de d e Escoci E scocia. a. Se dec´ dec´ıa que padec´ padec´ıa tuberculosis tuberc ulosis y Cardano C ardano dijo que la sab´ sab´ıa curar, lo cual no era cierto. Viaj´o a Edimburgo y afortunadamente para el obispo, y tambi´en en para Cardano, Cardano , result´o que la enfermedad era asma. Cuando Cardano pas´o por Londres en el viaje de vuelta, fue recibido por el joven rey Eduardo VI a quien hizo un hor´oscopo. oscopo. Le predijo larga vida y un pr´ospero ospero futuro, lo cual le puso en una situaci´on on inc´omoda omoda cuando el chico muri´ o poco despu´es.En es.En varias ocasiones, Cardano fue profesor de matem´aticas aticas de las universidade universidadess de Mil´an, an, Pavia y Bolonia, teniendo que dimitir de todas todas ellas ellas por alg´ un un esc´andalo andal o relacionado relacio nado con ´el. el. Cardano destaca por sus trabajos de ´algebra. algebra. En 1539 public´o su libro de aritm´etica etica Practica arithmetica et mensurandi singulares. Public´o las soluciones a las ecuaciones de tercer y cuarto grado en su Ars magna datado en 1545. La soluci´on on a un caso particular de ecuaci´on o n c´ ubica ubica x3 + ax = b
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reduciendo la ecuacion a: z 3 + pz + + q = = 0
donde:
a2
2a3 ab p = b − ; q = = − + c = 0 3 27 3 Haciendo nuevamente un cambio de variable, z=u+v, se tiene: z 3 = u 3 + v 3 + 3uv (u + v ) = u 3 + v 3 + 3uvz
Igualando con la ecuaci´on on cubica reducida se tiene que: − p = 3uv ; −q = = u 3 + v 3
Resolviendo el sistema se obtiene: 3
v =
El discriminante es:
−q ± ±
+ q 2
2
4 p3
4 p3 27
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3.1
PROG PROGRA RAMA MA CARD CARDAN ANO O
program CARDANO implicit implicit none DOUBLE PRECISION::o,o1,a, PRECISION::o,o1,a,a1,b,b1,c a1,b,b1,c,c1 ,c1 COMPLEX::p,q,dis,x !PARA !PARA UN POLINI POLINIMIO MIO DE LA FORMA FORMA !OX^3+AX^2+BX+C=0 ´ WRITE( WRITE(*,* *,*)’ )’ ESCRIB ESCRIBA A LOS LOS VALORE VALORES S DE LOS LOS COEFIC COEFICIEN IENTES TES DE LA ECUAC ECUACI I ON ON CUBICA’ CUBICA’ WRITE(*,*)’OX^3+AX^2+BX+C=0’ WRITE(*, WRITE(*,*)’* *)’***ES **ESCRIB CRIBA A O***’,’ O***’,’ ’,’DEBE ’,’DEBE DE SER DISTINTO DISTINTO DE CERO’ CERO’ READ*,o WRITE(*,*)’ WRITE(*,*)’ INGRESO’,o,’X^3’ INGRESO’,o,’X^3’ WRITE(*,*)’***ESCR WRITE(*,*)’***ESCRIBA IBA A***’ READ*,a WRITE(*,*)’ WRITE(*,*)’ INGRESO’,a,’X^2’ INGRESO’,a,’X^2’ WRITE(*,*)’***ESCR WRITE(*,*)’***ESCRIBA IBA B***’ READ*,b WRITE(*,*)’ WRITE(*,*)’ INGRESO’,b,’X’ INGRESO’,b,’X’
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elseif (o.EQ.1)THEN (o.EQ.1)THEN p=(3.0*b-a**2.0)/(3.0) write(*,*)’p’,p q=(2.0*a**3.0-9.0*a*b+27.0*c)/(27.0) write(*,*)’q’,q WRITE(*,*)’####################################’ dis=(q*0.5)**2.0+(p/3.0)**3.0 write(*,*)’dis’,dis x=(-q/2.0+sqrt(dis))**0.33333-(+q/2.0+sqrt(dis))**0.33333-(a/3.0) !FACT !FACTORI ORICE CE EL SIGNO SIGNO POR LA RAIZ RAIZ CUBICA CUBICA WRITE(*,*)’####################################’ WRITE(*,*)’LA WRITE(*,*)’LA SOLUCION ES:’,x end end if STOP end program program CARDANO CARDANO
4
BIBLIOGRAF´ IA