Question

Valendo ( 5 PONTOS)

A soma de dois números é 6 e o quadrado do maior menos o triplo do menor é 22.Determine esses números.


S = {(-8;14);(1;5)}

Answer 1

x + y = 6
x² - 3y = 22

x + y = 6
y = 6 - x

x² + 3y = 22
x² - 3(6 - x) = 22
x² - 18 + 3x = 22
x² + 3x - 18 - 22 = 0
x² + 3x - 40 = 0

x' + x" = - b/a = - 3
x'.x" = c/a = - 40

dois numeros que somando dar - 3
e multiplicando dar - 40

logicamente sao: - 8 e 5
pois,
- 8 + 5 = - 3
- 8.5 = - 40

x' + y' = 6
- 8 + y' = 6
y' = 6 + 8
y' = 14

S1 = {- 8,14}

x" + y" = 6
5 + y" = 6
y" = 6 - 5
y" = 1

S2 ={5,1}

Answer 2

$\large\begin{array}{l} \textsf{Sejam x e y os n\'umeros procurados, onde y \'e o maior deles:}\\\\ \mathsf{y>x\qquad(i)}\\\\\\ \textsf{De acordo com o enunciado, devemos ter}\\\\ \left\{\! \begin{array}{rrrrrc} \mathsf{x}&\!\!\!+\!\!\!&\mathsf{y}&\!\!=\!\!&\mathsf{6}&\quad\mathsf{(ii)}\\ \mathsf{y^2}&\!\!\!-\!\!\!&\mathsf{3x}&\!\!=\!\!&\mathsf{22}&\quad\mathsf{(iii)} \end{array} \right. \end{array}$


$\large\begin{array}{l} \textsf{Isole y na equa\c{c}\~ao (ii), e substitua na equa\c{c}\~ao (iii):}\\\\ \mathsf{y=6-x}\\\\\\ \mathsf{(6-x)^2-3x=22}\\\\ \mathsf{36-12x+x^2-3x=22}\\\\ \mathsf{x^2-15x+36-22=0}\\\\ \mathsf{x^2-15x+14=0} \end{array}$


$\large\begin{array}{l} \textsf{Vou resolver esta equa\c{c}\~ao usando completamento de quadrados.}\\\\ \textsf{Podemos reescrever o lado esquerdo assim:}\\\\ \mathsf{x^2-2\cdot \dfrac{15}{2}\,x+14=0}\\\\ \mathsf{x^2-2\cdot \dfrac{15}{2}\,x=-14} \end{array}$


$\large\begin{array}{l} \textsf{Com o objetivo de tornar o lado esquerdo um trin\^omio}\\\textsf{quadrado perfeito, adicionamos }\mathsf{\left(\frac{15}{2} \right)^{\!2}}\textsf{ aos dois lados da}\\\textsf{equa\c{c}\~ao:}\\\\ \mathsf{x^2-2\cdot \dfrac{15}{2}\,x+\left(\dfrac{15}{2}\right)^{\!2}=-14+\left(\dfrac{15}{2}\right)^{\!2}}\\\\ \mathsf{\left(x-\dfrac{15}{2}\right)^{\!2}=-14+\left(\dfrac{15}{2}\right)^{\!2}}\\\\ \mathsf{\left(x-\dfrac{15}{2}\right)^{\!2}=-14+\dfrac{225}{4}}\\\\ \mathsf{\left(x-\dfrac{15}{2}\right)^{\!2}=-\,\dfrac{56}{4}+\dfrac{225}{4}} \end{array}$

$\large\begin{array}{l} \mathsf{\left(x-\dfrac{15}{2}\right)^{\!2}=\dfrac{-56+225}{4}}\\\\ \mathsf{\left(x-\dfrac{15}{2}\right)^{\!2}=\dfrac{169}{4}}\\\\\\ \textsf{Tomando as ra\'izes quadradas, temos}\\\\ \mathsf{x-\dfrac{15}{2}=\pm\,\sqrt{\dfrac{169}{4}}}\\\\ \mathsf{x-\dfrac{15}{2}=\pm\,\sqrt{\dfrac{13^2}{2^2}}} \end{array}$

$\large\begin{array}{l} \mathsf{x-\dfrac{15}{2}=\pm\,\sqrt{\left(\dfrac{13}{2}\right)^{\!\!2}}}\\\\ \mathsf{x-\dfrac{15}{2}=\pm\,\dfrac{13}{2}}\\\\ \mathsf{x=\pm\,\dfrac{13}{2}+\dfrac{15}{2}}\\\\ \mathsf{x=\dfrac{\pm\,13+15}{2}} \end{array}$


$\large\begin{array}{l} \begin{array}{rcl} \mathsf{x=\dfrac{-13+15}{2}}&~\textsf{ ou }~&\mathsf{x=\dfrac{13+15}{2}}\\\\ \mathsf{x=\dfrac{2}{2}}&~\textsf{ ou }~&\mathsf{x=\dfrac{28}{2}}\\\\ \mathsf{x=1}&~\textsf{ ou }~&\mathsf{x=14} \end{array} \end{array}$


$\large\begin{array}{l} \bullet~~\textsf{Para }\mathsf{x=1,}\textsf{ encontramos}\\\\ \mathsf{y=6-1}\\\\ \mathsf{y=5\qquad\checkmark\qquad}\textsf{(satisfaz }\mathsf{y>x}\textsf{)}\\\\\\ \bullet~~\textsf{Para }\mathsf{x=14,}\textsf{ encontramos}\\\\ \mathsf{y=6-14}\\\\ \mathsf{y=-8\qquad\diagup\!\!\!\!\!\diagdown\qquad}\textsf{(n\~ao satisfaz }\mathsf{y>x}\textsf{)} \end{array}$


$\large\begin{array}{l} \textsf{Portanto, a \'unica solu\c{c}\~ao para o problema proposto \'e}\\\\ \boxed{\begin{array}{c}\mathsf{(x,\,y)=(1,\,5)} \end{array}}\\\\\\ \textsf{Conjunto solu\c{c}\~ao: }\mathsf{S=\big\{(1,\,5)\big\}} \end{array}$


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


Tags: sistema equação segundo grau quadrática soma diferença completamento de quadrados substituição