Question

Dada a função: $f(x)=ax^2+bx+c$, temos que:

$\bullet\ {a} \neq 0\ ;\\\bullet{a,b\ e\ c\ s\~ao\ n\'umeros\ reais;$
$\bullet\ f(x)\ possui\ duas\ ra\'izes\ reais\ iguais;\ e$
$\bullet\ O\ gr\'afico\ da\ f(x)\ passa\ pelos\ pontos\ (-1,9)\ e\ (3,1).$

Assim sendo, calcule o valor de $a+b+c$

Answer 1

Olá,

A forma fatorada da equação é:

$F(x)=a\cdot (x-x_{1})\cdot (x-x_{2})\quad\quad\quad\quad*x_{1}=x_{2}=R\\\\F(x)=a\cdot (x-R)\cdot (x-R)\\\\F(x)=a\cdot (x-R)^2$

$\text{Substituindo f(x) e x pelo dos pontos que foram dados:}\\\\\text{-Para o primeiro ponto:}\\\\F(x)=a\cdot (x-R)^2\\\\9=a\cdot (-1-R)^2\\\\a=\dfrac{9}{(-1-R)^2}\\\\\text{-Para o segundo ponto:}\\\\F(x)=a\cdot (x-R)^2\\\\1=a\cdot (3-R)^2\\\\a=\dfrac{1}{(3-R)^2}$

$\text{Para descobrir "R" vamos igualar as duas equac\~oes:}\\\\\dfrac{9}{(-1-R)^2}=\dfrac{1}{(3-R)^2}\\\\9\cdot (3-R)^2=(-1-R)^2\\\\9\cdot (9-6R+R^2)=1+2R+R^2\\\\81-54R+9R^2=1+2R+R^2\\\\8R^2-56R+80=0$

$8R^2-56R+80=0\\\\S=\dfrac{-b}{a}\\\\S=\dfrac{-(-56)}{8}\\\\S=7\\\\P=\dfrac{c}{a}\\\\P=\dfrac{80}{8}\\\\P=10\\\\\boxed{\text{As ra\'izes s\~ao 5 ou 2}}$

$\text{Se a ra\'iz for 2 "a" ser\'a:}\\\\a=\dfrac{1}{(3-R)^2}\\\\a=\dfrac{1}{(3-2)^2}\\\\a=\dfrac{1}{(1)^2}\\\\a=1\\\\\text{E a lei da func\~ao ficar\'a:}\\\\f(x)=a\cdot (x-R)^2\\\\f(x)=1\cdot (x-2)^2\\\\f(x)=(x-2)\cdot (x-2)\\\\f(x)=x^2-4x+4\\\\a=1\quad\quad b=-4\quad\quad c=4\\\\a+b+c=1$

$\text{Para R=5 temos que:}\\\\a=\frac{1}{(3-R)^2}\\\\a=\frac{1}{(3-5)^2}\\\\a=\dfrac{1}{(-2)^2}\\\\a=\dfrac{1}{4}\\\\\text{Logo, a lei de forma\c{c}\~ao fica:}\\\\f(x)=a\cdot (x-R)^2\\\\f(x)=\frac{1}{4}\cdot (x^2-10x+25)\\\\f(x)=\frac{1}{4}x^2-\frac{10}{4}x+\frac{16}{4}\\\\a+b+c=\dfrac{36}{4}\\\\a+b+c=4$

a+b+c pode ser 1 ou 4.

Até mais!

Answer 2

$\large\begin{array}{l} \textsf{\'E dada a seguinte fun\c{c}\~ao do segundo grau:}\\\\ \mathsf{f(x)=ax^2+bx+c\qquad(a,\,b,\,c\in\mathbb{R},\,a\ne 0)}\\\\\\ \textsf{Temos a informa\c{c}\~ao de que f possui duas ra\'izes reais iguais.}\\\textsf{Vamos chamar esta raiz por r.}\\\\\\ \textsf{Ent\~ao, a lei de f pode ser escrita na forma}\\\\ \mathsf{f(x)=a(x-r)^2\qquad(i)}\\\\ \textsf{pois r \'e uma raiz dupla.} \end{array}$


$\large\begin{array}{l} \textsf{Usando o que foi fornecido pelo enunciado.}\\\\ \bullet~~\textsf{O gr\'afico de f passa pelo ponto }\mathsf{(-1,\,9):}\\\\ \mathsf{f(-1)=9}\\\\ \mathsf{a(-1-r)^2=9\qquad(ii)}\\\\\\ \bullet~~\textsf{O gr\'afico de f passa pelo ponto }\mathsf{(3,\,1):}\\\\ \mathsf{f(3)=1}\\\\ \mathsf{a(3-r)^2=1\qquad(iii)} \end{array}$


$\large\begin{array}{l} \textsf{Como }\mathsf{a\ne 0,}\textsf{ deve-se ter necessariamente}\\\\ \begin{array}{rcl} \mathsf{-1-r\ne 0}&~\textsf{ e }~&\mathsf{3-r\ne 0}\\\\ \mathsf{r\ne -1}&~\textsf{ e }~&\mathsf{r\ne 3} \end{array}\\\\\\ \textsf{Ent\~ao podemos isolar a vari\'avel a nas equa\c{c}\~oes (ii) e (iii):}\\\\ \mathsf{a=\dfrac{9}{(-1-r)^2}\qquad(iv)}\\\\ \mathsf{a=\dfrac{1}{(3-r)^2}\qquad(v)} \end{array}$


$\large\begin{array}{l} \textsf{Igualando (iv) e (v),}\\\\ \mathsf{\dfrac{9}{(-1-r)^2}=\dfrac{1}{(3-r)^2}}\\\\ \mathsf{9\cdot (3-r)^2=1\cdot (-1-r)^2}\\\\ \mathsf{9\cdot (9-6r+r^2)=1+2r+r^2}\\\\ \mathsf{81-54r+9r^2=1+2r+r^2}\\\\ \mathsf{81-54r+9r^2-1-2r-r^2=0}\\\\ \mathsf{9r-r^2-54r-2r+81-1=0} \end{array}$

$\large\begin{array}{l} \mathsf{8r^2-56r+80=0}\\\\ \mathsf{8\cdot (r^2-7r+10)=0}\\\\ \mathsf{r^2-7r+10=0} \end{array}$


$\large\begin{array}{l} \textsf{Poder\'iamos resolver usando B\'ascara, mas vou fatorar o lado}\\\textsf{esquerdo por agrupamento.}\\\\ \textsf{Reescrevendo convenientemente }\mathsf{-7r}\textsf{ como }\mathsf{-2r-5r:}\\\\ \mathsf{r^2-2r-5r+10=0}\\\\ \mathsf{r(r-2)-5(r-2)=0}\\\\ \mathsf{(r-2)(r-5)=0}\\\\ \begin{array}{rcl} \mathsf{r-2=0}&~\textsf{ ou }~&\mathsf{r-5=0}\\\\ \mathsf{r=2}&~\textsf{ ou }~&\mathsf{r=5} \end{array} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{Para }\mathsf{r=2,}\textsf{ encontramos}\\\\ \mathsf{a=\dfrac{9}{(-1-2)^2}}\\\\ \mathsf{a=\dfrac{9}{(-3)^2}}\\\\ \mathsf{a=\dfrac{9}{9}}\\\\ \mathsf{a=1}\\\\\\ \textsf{a lei da fun\c{c}\~ao f \'e}\\\\ \mathsf{f(x)=1\cdot (x-2)^2}\\\\ \boxed{\begin{array}{c}\mathsf{f(x)=x^2-4x+4}\end{array}}\quad\Rightarrow\quad\mathsf{a=1,\,b=-4,\,c=4} \end{array}$


$\large\begin{array}{l} \textsf{e a soma pedida ser\'a}\\\\ \mathsf{a+b+c}\\\\ =\mathsf{1+(-4)+4}\\\\ =\mathsf{1\qquad\checkmark} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{Para }\mathsf{r=5,}\textsf{ encontramos}\\\\ \mathsf{a=\dfrac{9}{(-1-5)^2}}\\\\ \mathsf{a=\dfrac{9}{(-6)^2}}\\\\ \mathsf{a=\dfrac{9}{36}}\\\\ \mathsf{a=\dfrac{\diagup\!\!\!\! 9}{\diagup\!\!\!\! 9\cdot 4}}\\\\ \mathsf{a=\dfrac{1}{4}}\\\\\\ \textsf{a lei da fun\c{c}\~ao f \'e}\\\\ \mathsf{f(x)=\frac{1}{4}\cdot (x-5)^2}\end{array}$

$\large\begin{array}{l} \mathsf{f(x)=\frac{1}{4}\cdot (x^2-10x+25)}\\\\ \boxed{\begin{array}{c}\mathsf{f(x)=\frac{1}{4}\,x^2-\frac{10}{4}\,x+\frac{25}{4}} \end{array}}\quad\Rightarrow\quad\mathsf{a=\frac{1}{4},\,b=-\frac{10}{4},\,c=\frac{25}{4}}\\\\\\ \textsf{e a soma pedida ser\'a}\\\\ \mathsf{a+b+c}\\\\ =\mathsf{\frac{1}{4}-\frac{10}{4}+\frac{25}{4}}\\\\ =\mathsf{\frac{1-10+25}{4}}\\\\ =\mathsf{\frac{16}{4}}\\\\ =\mathsf{4\qquad\checkmark} \end{array}$


$\large\begin{array}{l} \textsf{Logo, o problema proposto admite duas solu\c{c}\~oes.} \end{array}$


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$\large\begin{array}{l} \textsf{D\'uvidas comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$


Tags: desafio função quadrática segundo grau gráfico calcular soma coeficiente raiz sistema equação fatoração por agrupamento Báscara