Question

03) Determine o conjunto solução das equações do segundo grau na incógnita x.
a) x2 - 7x + 10 = 0
d) 4x2 - 20 = 0
g) 6x2 + x - 1 = 0
b) x2 - 8x + 12 = 0
e) -5x2 + 4x = 0
h) 4x2 + 9 = 12x
c) x2 + 2x - 8 = 0
f) 2x2 - 10x + 16 = 0
i) (x - 5)2 = 1
mim ajudem preciso muito​

Answer 1

Resolvendo as equações do 2º grau, temos como conjunto solução (considerando U = ℝ):

a) S = {2 ; 5}; b) S = {2 ; 6}; c) S = {– 4 ; 2}; d) S = {– √5 ; √5}; e) S = {0 ; 4/5}; f) S = ∅; g) S = {– 1/2 ; 1/3}; h) S = {3/2}; i) S = {4 ; 6}.

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Desejamos determinar o conjunto solução S de cada equação quadrática listadas de a) até i). Obs.: eu coloquei os itens na ordem alfabética apesar de estar bagunçado na tarefa.

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Consideremos uma equação do 2º grau completa na forma ax² + bx + c = 0, com a, b e c ∈ ℝ e a ≠ 0. Logo uma equação incompleta: com b = 0 é ax² + c = 0; com c = 0 é ax² + bx = 0. Assim, temos três possíveis formas de resolução,

• podemos resolver pela fórmula de Bhaskara, que consiste em calcular o delta (Δ = b² – 4ac), e depois partir pra formula x = (–b ± √Δ)/2a. Detalhe, se Δ > 0 então x₁ e x₂ ∈ ℝ | x₁ ≠ x₂, se Δ = 0 então x₁ e x₂ ∈ ℝ | x₁ = x₂, e se Δ < 0 então x₁ e x₂ ∉ ℝ;
• podemos resolver por fatoração (agrupamento), quando: for possível desmembrar o segundo termo da eq. completa numa soma a fim de colocar em evidência com os outros termos. Também em caso de eq. incompleta com c = 0 podemos colocar o fator comum em evidência;
• podemos resolver semelhante à uma eq. do 1º grau, quando: tivermos uma eq. incompleta com b = 0, assim podemos isolar x e extrair a raiz quadrada de ambos os membros.

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Com tudo em mente, acompanhe:

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Item a)

$\large\begin{array}{l}\sf\implies~~~x^2-7x+10=0\\\\\sf\iff~~~x^2-2x-5x+10=0\\\\\sf\iff~~~x(x-2)-5(x-2)=0\\\\\sf\iff~~~(x-2)(x-5)=0\\\\\iff~~\begin{cases}\sf x-2=0\\\sf x-5=0\end{cases}\\\\\iff~~\begin{cases}\sf x_1=2\\\sf x_2=5\end{cases}\\\\\quad\!\therefore\quad~~\,\boldsymbol{\boxed{\sf S=\big\{2~~;~~5\big\}}}\end{array}$

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Item b)

$\large\begin{array}{l}\sf\implies~~~x^2-8x+12=0\\\\\sf\iff~~~x^2-2x-6x+12=0\\\\\sf\iff~~~x(x-2)-6(x-2)=0\\\\\sf\iff~~~(x-2)(x-6)=0\\\\\iff~~\begin{cases}\sf x-2=0\\\sf x-6=0\end{cases}\\\\\iff~~\begin{cases}\sf x_1=2\\\sf x_2=6\end{cases}\\\\\quad\!\therefore\quad~~\,\boldsymbol{\boxed{\sf S=\big\{2~~;~~6\big\}}}\end{array}$

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Item c)

$\large\begin{array}{l}\sf\implies~~~x^2+2x-8=0\\\\\sf\iff~~~x^2+4x-2x-8=0\\\\\sf\iff~~~x(x+4)-2(x+4)=0\\\\\sf\iff~~~(x+4)(x-2)=0\\\\\iff~~\begin{cases}\sf x+4=0\\\sf x-2=0\end{cases}\\\\\iff~~\begin{cases}\sf x_1=-\,4\\\sf x_2=2\end{cases}\\\\\quad\!\therefore\quad~~\,\boldsymbol{\boxed{\sf S=\big\{\!\!-4~~;~~2\big\}}}\end{array}$

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Item d)

$\large\begin{array}{l}\sf\implies~~~4x^2-20=0\\\\\sf\iff~~~4x^2=20\\\\\sf\iff~~~x^2=\dfrac{20}{4}\\\\\sf\iff~~~x^2=5\\\\\sf\iff~~~\sqrt{x^2}=\sqrt{5}\\\\\sf\iff~~~|x|=\sqrt{5}\\\\\sf\iff~~~x=\pm\:\sqrt{5}\\\\\sf\iff~~~x_1=-\,\sqrt{5}~~\vee~~x_2=\sqrt{5}\\\\\quad\!\therefore\quad~~\,\boldsymbol{\boxed{\sf S=\Big\{\!\!-\sqrt{5}~~;~~\sqrt{5}\Big\}}}\end{array}$

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Item e)

$\large\begin{array}{l}\sf\implies~~\,-5x^2+4x=0\\\\\sf\iff~~~x(-\,5x+4)=0\\\\\iff~~\begin{cases}\sf x=0\\\sf-\,5x+4=0\end{cases}\\\\\iff~~\begin{cases}\sf x_1=0\\\\\sf x_2=\dfrac{4}{5}\end{cases}\\\\\quad\!\therefore\quad~~\,\boldsymbol{\boxed{\sf S=\bigg\{0~~;~~\dfrac{4}{5}\bigg\}}}\end{array}$

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Item f)

$\large\begin{array}{l}\sf\implies~~~2x^2-10x+16=0\\\\\sf\iff~~~2(x^2-5x+8)=0\\\\\sf\iff~~~x^2-5x+8=2\cdot0\\\\\sf\iff~~~x^2-5x+8=0\end{array}$

Neste caso não é possível fatorar, então vamos pela fórmula de Bhaskara mesmo:

$\large\begin{array}{l}\sf\implies~~~\Delta=b^2-4ac\\\\\sf\iff~~~\Delta=(-5)^2-4\cdot1\cdot8\\\\\sf\iff~~~\Delta=25-32\\\\\sf\iff~~~\Delta=-\,7\end{array}$

Veja que Δ < 0, ∴ x₁ e x₂ ∉ ℝ. Assim o conjunto solução é vazio:

             $\large\begin{array}{l}\sf\boldsymbol{\boxed{\sf S=\Large\varnothing}}\end{array}$

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Item g)

$\large\begin{array}{l}\sf\implies~~~6x^2+x-1=0\\\\\sf\iff~~~6x^2+3x-2x-1=0\\\\\sf\iff~~~3x(2x+1)-1(2x+1)=0\\\\\sf\iff~~~(2x+1)(3x-1)=0\\\\\iff~~\begin{cases}\sf2x+1=0\\\sf3x-1=0\end{cases}\\\\\iff~~\begin{cases}\sf x_1=-\dfrac{1}{2}\\\\\sf x_2=\dfrac{1}{3}\end{cases}\\\\\quad\!\therefore\quad~~\,\boldsymbol{\boxed{\sf S=\bigg\{\!\!-\dfrac{1}{2}~~;~~\dfrac{1}{3}\bigg\}}}\end{array}$

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Item h)

$\large\begin{array}{l}\sf\implies~~~4x^2+9=12x\\\\\sf\iff~~~4x^2-12x+9=0\end{array}$

Nesse caso temos um quadrado perfeito [a² – 2ab + b² = (a – b)²]:

$\large\begin{array}{l}\!\!\sf\iff~~~(2x)^2-2\cdot2\cdot3+(3)^2=0\\\\\sf\iff~~~(2x-3)^2=0\\\\\sf\iff~~~\sqrt{(2x-3)^2}=\sqrt{0}\\\\\sf\iff~~~|2x-3|=0\\\\\sf\iff~~~2x-3=0\\\\\sf\iff~~~x_1=x_2=\dfrac{3}{2}\\\\\quad\!\therefore\quad~~\,\boldsymbol{\boxed{\sf S=\bigg\{\dfrac{3}{2}\bigg\}}}\end{array}$

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Item i)

$\large\begin{array}{l}\sf\implies~~~(x-5)^2=1\\\\\sf\iff~~~\sqrt{(x-5)^2}=\sqrt{1}\\\\\sf\iff~~~|x-5|=1\\\\\sf\iff~~~x-5=\pm\:1\\\\\iff~~\begin{cases}\sf x-5=1\\\sf x-5=-\,1\end{cases}\\\\\iff~~\begin{cases}\sf x_1=6\\\sf x_2=4\end{cases}\\\\\quad\!\therefore\quad~~\,\boldsymbol{\boxed{\sf S=\big\{4~~;~~6\big\}}}\end{array}$

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Answer 2

Resposta:

Explicação passo a passo:

Respostas:

A) S = x¹= 5 , x²= 2

B) S= x¹= 6, x² = 2

C) S= x¹= 2 , x²= -4

D) S = x¹= +√20/4 , x²= -√20/4

E) S= x¹=o, x²= -9/4

F) S= x¹ =32, x²= -27

G) S= x¹ =1/3, x²= 1/2

H) S= x¹= -9/4, x²= 9/4

I) S= x¹= 6 , x²= 4

Explicação passo a passo:

A)

X= -b±√b²-4×a×c

____________

2×a

X= -(-7)±√7²-4×1×10

_____________

2

X= 7±√49-4×1×10

__________

2

X= 7±√9

_____

2

X= 7±3

_____

2

X¹= 7+3 = 10 = 5

___ ___

2 2

X²= 7-3= 4 = 2

___ ___

2 2

B) x= -(-8)±√(-8)²-4×1×12

_______________

2

X= 8±√64-4×1×12

___________

2

X=8±√16

______

2

X= 8±4

____

2

X¹= 8+4 = 12 = 6

____ ___

2 2

X²= 8-4 = 4 = 2

___ ___

2 2

C)

X= -2±√2²-4×1×-8

____________

2

X= -2±√36

________

2

X= -2±6

_____

2

X¹= -2+6 = 4 = 2

____ ____

2 2

X²= -2-6 = -8 = -4

____ ____

2 2

D)

4x² =20

X²= 20/4

x= ±√20/4

E)

-5x²+4x=0

X¹=0

X²= -5/4

F)

X=-(-10)±√-10²-4×2×16

______________

4

X= 10±√-100-4×2×16

____________

4

X=10±-118

______

4

X¹= 10-118 = -108 = -27

____ _____

4 4

X²= 10+118 = 128 = 32

____ ____

4 4

G)

x= -1± √1²-4×6×-1

___________

2×6

X= -1±√25

______

12

X= -1±√5

______

12

X¹= -1+5

____

12

= 4

___

12

= 1

__

3

X²= -1-5

___

12

= -6

___

12

= -1

___

2

H)

4x²+9=12x

4x²=-9

X²= -9/4

X= ±-9/4

I)

(x-5)²=1

√(x-5)²=√1

[ X-5]=1

X-5=±1

x-5=1

X-5=-1

X¹= 6

X²=4