Exercício de Questão da Unifesp
Exercício:
Certo dia um professor de matematica desafiou seus alunos a descobrirem as idades x,y,z, em anos, de seus tres filhos,dizendo ser o produto delas igual a 40.
De pronto,os alunos protestaram:a informação “x.y.z=40” era insuficiente para uam resposta correta,em vista de terem encontrado 6 ternas de fatores do número 40 cujo produto é 40.O professor concordou e disse,apontando para um dos alunos,que a soma x+y+z das idades (em anos) era igual ao numero que se podia ver estampado na camisa que ele estava usando.Minutos deposi os alunos disseram continuar impossivel responder com segurança,mesmo sabendo que a soma era um número conhecido,o que levou o professor a perceber que eles racionavam corretamente (chegando a um impasse,provocado por duas ternas)
Satisfeito,o professor acrescentou então duas informações definitivas:seus tres filhos haviam nascido no mesmo mês e naquele exato dia,o caçula estava fazendo aniversario. Neste caso a resposta correta é:
a)1,5,8
b)1,2,20
c)1,4,10
d)1,1,40
e)2,4,5
Informações Adicionais:
Obrigada desde já pra quem responder
O problema fala que são 6 ternas de fatores de 40. mas a questão dá apenas 5 alternativas. Qual seria a 6ª terna ?
(2 , 2 , 10)
Note que somando as idades nas 6 ternas, duas somas são iguais (1+5+8 = 2+2+10 = 14). Os alunos ficaram na dúvida entre elas, pois se fosse uma das outras ternas, quando eles viram o número estampado na camisa do colega, eles teriam descoberto qual a verdadeira.
Aí o professor disse que o caçula estava fazendo aniversário naquele dia. Logo, os filhos mais novos não são gêmeos.
Portanto, as idades dos filhos são 1, 5 e 8 anos.
4x²-3x-1=0
sendo: a= 4, b= -3 e c= -1
Δ= b²-4.a.c
Δ = (-3)² – 4.4.(-1)
Δ = 9 – 16 .(-1)
Δ = 9 + 16
logo Δ = 25
fórmula de bhaskara:
-b ± √Δ / 2.a
agora substitui :
3 ± √25 / 2.4
sendo √25 = 5
e 2.4= 8
fica :
x¹ = 3 – 5 / 8 = – 2/8 simplificado por 2 = -1/8 ou seja: -8
x² = 3+ 5 / 8 = 8/8 = 1
S = {1, -8}.
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