Matéria: Matemática
Autor: ed704184
Perguntado 5 anos atrás

Resolva as equações de segundo grau completas, utilizando a Fórmula de Bhaskara:

: 2x 2 + 7x + 5 = 0

: x2 + 20x − 525 = 0

x2 − 16x + 64 = 0

x2 − 2x + 2 = 0

me ajudem pfvvv

Respostas

respondido por: B0Aventura

Resposta:

$2x^{2} +7x+5=0\\ \\ coeficientes:~a=2,~b=7,~c=5\\ \\ x=\frac{-b+-\sqrt{b^{2}-4.a.c } }{2.a} \\ \\ x=\frac{-7+-\sqrt{49-4.2.5} }{4} \\ \\ x=\frac{-7+-\sqrt{49-40} }{4} \\ \\ x=\frac{-7+-\sqrt{9} }{4} \\ \\ x'=\frac{-7-3}{4} =\frac{-10}{4} =-\frac{5}{2} =-2,5\\ \\ x''=\frac{-7+3}{4} =\frac{-4}{4} =-1\\ \\ \left \{ {{x'=-2,5;~~x''=-1} {}} \}$

$x^{2} +20x-525=0\\ \\ coeficientes:~a=1,~b=20,~c=525\\ \\ x=\frac{-b+-\sqrt{b^{2}-4.a.c } }{2.a} \\ \\ x=\frac{-20+-\sqrt{20^{2}-4.1.525 } }{2.1} \\ \\ x=\frac{-20+-\sqrt{400-2100} }{2} \\ \\ x=\frac{-20+-\sqrt{-1700} }{2} \\ \\ raiz~quadrada~de~numero~negativo:~nao~existe~no~conjunto~real$

$x^{2} -16x+64=0\\ \\ coeficientes:~a=1,~b=-16,~c=64\\ \\ x=\frac{-b+-\sqrt{(-16)^{2}-4.1.64 } }{2.1} \\ \\ x=\frac{-(-16)+-\sqrt{256-4.1.64} }{2} \\ \\ x=\frac{16+-\sqrt{256-256} }{2} \\ \\ x=\frac{16+-\sqrt{0} }{2} \\ \\ x=\frac{16}{2} \\ \\ x=8$

$x^{2} -2x+2=0\\ \\ coeficientes:~a=1,~b=-2,~c=2\\ \\ x=\frac{-b+-\sqrt{b^{2}-4.a.c } }{2.1} \\ \\ x=\frac{-(-2)+-\sqrt{(-2)^{2} -4.1.2} }{2} \\ \\ x=\frac{2+-\sqrt{4-8} }{2} \\ \\ x=\frac{2+-\sqrt{-4} }{2} \\ \\ raiz~quadrada~de~numero~negativo:~nao~pertence~ao~conjunto~dos~numeros~reais$

respondido por: Nasgovaskov

Temos equações do 2º grau, pois se encontram na forma ax² + bx + c = 0, onde os coeficientes são:

''a'' que multiplica x²
''b'' que multiplica x
''c'' que é o termo independente

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Devemos encontrar as raízes das equações pela fórmula de Bhaskara, consistindo na fórmula:

$\boxed{\begin{array}{l}\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\:,~onde~\Delta=b^2-4ac\\\\\end{array}}$

Dessa forma, podemos calcular o delta primeiro, e seguir para a formula da determinação de x. Assim, basta identificar os coeficientes das equações, e substituir na fórmula.

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* obs.: quando Δ > 0, a equação admite duas raízes reais e diferentes. Quando Δ = 0, a equação admite duas raízes reais e iguais. Quando Δ < 0 a equação não admite raízes reais.

( I )

$\begin{array}{l}\sf2x^2+7x+5=0\end{array}$

a = 2
b = 7
c = 5

$\begin{array}{l}\sf\Delta=b^2-4ac\\\\\sf\Delta=(7)^2-4\cdot2\cdot5\\\\\sf\Delta=49-40\\\\\sf\Delta=9~~\to~\Delta > 0~\therefore~x'\,\neq\,x''\in\mathbb{R}\\\sf \\\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(7)\pm\sqrt{9}}{2\cdot2}~\Rightarrow~\sf x=\dfrac{-7\pm3}{4}\\\\\sf x=\begin{cases}\sf x'=\dfrac{-7+3}{4}=-\dfrac{4}{4}=-1\\\\\sf x''=\dfrac{-7-3}{4}=-\dfrac{10}{4}=-\dfrac{5}{2}\end{cases}\end{array}$

Conjunto solução...

$\boxed{\begin{array}{l}\sf S=\bigg\{\!-1~~;~-\dfrac{5}{2}\bigg\}\end{array}}$

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( II )

$\begin{array}{l}\sf x^2+20x-525=0\end{array}$

a = 1
b = 20
c = - 525

$\begin{array}{l}\sf\Delta=b^2-4ac\\\\\sf\Delta=(20)^2-4\cdot1\cdot(-525)\\\\\sf\Delta=400+2100\\\\\sf\Delta=2500~~\to~\Delta > 0~\therefore~x'\,\neq\,x''\in\mathbb{R}\\\sf \\\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(20)\pm\sqrt{2500}}{2\cdot1}~\Rightarrow~\sf x=\dfrac{-20\pm50}{2}\\\\\sf x=\begin{cases}\sf x'=\dfrac{-20+50}{2}=\dfrac{30}{2}=15\\\\\sf x''=\dfrac{-20-50}{2}=-\dfrac{70}{2}=-35\end{cases}\end{array}$

Conjunto solução...

$\boxed{\begin{array}{l}\sf S=\Big\{\!-35~~;~~15\Big\}\end{array}}$

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( III )

$\begin{array}{l}\sf x^2-16x+64=0\end{array}$

a = 1
b = - 16
c = 64

$\begin{array}{l}\sf\Delta=b^2-4ac\\\\\sf\Delta=(-16)^2-4\cdot1\cdot64\\\\\sf\Delta=256-256\\\\\sf\Delta=0~~\to~~\therefore~x'=x''\in\mathbb{R}\\\sf \\\\\sf x=\dfrac{-b\pm\sqrt{\Delta}}{2a}\\\\\sf x=\dfrac{-(-16)\pm\sqrt{0}}{2\cdot1}~\Rightarrow~\sf x=\dfrac{16\pm0}{2}\\\\\sf x=\begin{cases}\sf x'=x''=\dfrac{16}{2}=8\end{cases}\end{array}$

Conjunto solução...

$\boxed{\begin{array}{l}\sf S=\Big\{\:8\:\Big\}\end{array}}$

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( IV )

$\begin{array}{l}\sf x^2-2x+2=0\end{array}$

a = 1
b = - 2
c = 2

$\begin{array}{l}\sf\Delta=b^2-4ac\\\\\sf\Delta=(-2)^2-4\cdot1\cdot2\\\\\sf\Delta=4-8\\\\\sf\Delta=-4~~\to~\Delta < 0~\therefore~x\notin\mathbb{R}\end{array}$