Question

Preencha a tabela identificando os coeficientes, os discriminante e as raízes das seguintes funções polinominais do segundo grau. Me ajudem pfvr

Answer 1

Essa questão é sobre equações do segundo grau. As equações do segundo grau são representadas por ax² + bx + c = 0, onde a, b e c são os coeficientes da equação. Para encontrar as raízes dessas equações, devemos utilizar a fórmula de Bhaskara, dada por:

x = [-b ±√(b²-4ac)]/2a

• f(x) = x² + 2x

Os coeficientes são:

a = 1

b = 2

c = 0

O discriminante é:

Δ = 2² - 4·1·0

Δ = 4

As raízes são:

x = [-2 ±√4]/2·1

x = [-2 ± 2]/2

x' = 0

x'' = -2

• f(x) = x² - 13x + 40

Os coeficientes são:

a = 1

b = -13

c = 40

O discriminante é:

Δ = (-13)² - 4·1·40

Δ = 9

As raízes são:

x = [13 ±√9]/2·1

x = [13 ± 3]/2

x' = 8

x'' = 5

• f(x) = 2x² - 7x + 3

Os coeficientes são:

a = 2

b = -7

c = 3

O discriminante é:

Δ = (-7)² - 4·2·3

Δ = 25

As raízes são:

x = [7 ±√25]/2·2

x = [7 ± 5]/4

x' = 3

x'' = 1/2

• f(x) = x² - 2x - 3

Os coeficientes são:

a = 1

b = -2

c = -3

O discriminante é:

Δ = (-2)² - 4·1·(-3)

Δ = 16

As raízes são:

x = [2 ±√16]/2·1

x = [2 ± 4]/2

x' = 3

x'' = -1

Answer 2

A tabela preenchida com os coeficientes, discriminantes e raízes encontra-se logo abaixo no campo "Resposta".

Acompanhe a solução:

• Numa equação de 2º grau padrão, temos: ax²+bx+c=0.
• Para calcular as raízes, basta aplicar a fórmula de Bháskara: $\boxed{\dfrac{-b\pm\sqrt{\Delta} }{2\cdot a}}$
• O discriminante (Δ) é dado por: $\boxed{\Delta=b^2-4\cdot a\cdot c}$.

Com isto, vamos aos cálculos!

Cálculo:

>>> 1ª equação:

→ Identificando os coeficientes:

$\large\begin {array}{l}\Large\boxed{\boxed{a=1, b= 2, c=0}}\Huge\checkmark\end {array}$

→ discriminante (Δ):

$\large\begin {array}{l}\Delta=b^2-4\cdot a\cdot c\\\\\Delta=2^2-4\cdot 1\cdot 0\\\\\Delta=2^2-0\\\\\Large\boxed{\boxed{\Delta=4}}\Huge\checkmark\end {array}$

→ Raízes:

$\large\begin {array}{l}x=\dfrac{-b\pm\sqrt{\Delta} }{2\cdot a}=\dfrac{-2\pm\sqrt{4} }{2\cdot 1}=\dfrac{-2\pm2 }{2}=\end {array}$

$\boxed{\large\begin {array}{l}x'=\dfrac{-2+2}{2}\\\\x'=\dfrac{0}{2}\\\\\Large\boxed{\boxed{x'=\;0\;}}\Huge\checkmark\;\end {array}}\quad\quad \quad \boxed{\large\begin {array}{l}x"=\dfrac{-2-2}{2}\\\\x"=\dfrac{-4}{2}\\\\\Large\boxed{\boxed{x"=-2}}\Huge\checkmark\end {array}}$

>>> 2ª equação:

→ Identificando os coeficientes:

$\large\begin {array}{l}\Large\boxed{\boxed{a=1, b= -13, c=40}}\Huge\checkmark\end {array}$

→ discriminante (Δ):

$\large\begin {array}{l}\Delta=b^2-4\cdot a\cdot c\\\\\Delta=(-13)^2-4\cdot 1\cdot 40\\\\\Delta=169-160\\\\\Large\boxed{\boxed{\Delta=9}}\Huge\checkmark\end {array}$

→ Raízes:

$\large\begin {array}{l}x=\dfrac{-b\pm\sqrt{\Delta} }{2\cdot a}=\dfrac{-(-13)\pm\sqrt{9} }{2\cdot 1}=\dfrac{13\pm3 }{2}=\end {array}$

$\boxed{\large\begin {array}{l}x'=\dfrac{13+3}{2}\\\\x'=\dfrac{16}{2}\\\\\Large\boxed{\boxed{x'=8}}\Huge\checkmark\;\end {array}}\quad\quad \quad \boxed{\large\begin {array}{l}x"=\dfrac{13-3}{2}\\\\x"=\dfrac{10}{2}\\\\\Large\boxed{\boxed{x"=5}}\Huge\checkmark\end {array}}$

>>> 3ª equação:

→ Identificando os coeficientes:

$\large\begin {array}{l}\Large\boxed{\boxed{a=2, b=-7, c=3}}\Huge\checkmark\end {array}$

→ discriminante (Δ):

$\large\begin {array}{l}\Delta=b^2-4\cdot a\cdot c\\\\\Delta=(-7)^2-4\cdot 2\cdot 3\\\\\Delta=49-24\\\\\Large\boxed{\boxed{\Delta=25}}\Huge\checkmark\end {array}$

→ Raízes:

$\large\begin {array}{l}x=\dfrac{-b\pm\sqrt{\Delta} }{2\cdot a}=\dfrac{-(-7)\pm\sqrt{25} }{2\cdot 2}=\dfrac{7\pm5 }{4}=\end {array}$

$\boxed{\large\begin {array}{l}x'=\dfrac{7+5}{4}\\\\x'=\dfrac{12}{4}\\\\\Large\boxed{\boxed{x'=\;3\;\;}}\Huge\checkmark\;\end {array}}\quad\quad \quad \boxed{\large\begin {array}{l}x"=\dfrac{7-5}{4}\\\\x"=\dfrac{2}{4}\\\\\Large\boxed{\boxed{x"=0,5}}\Huge\checkmark\end {array}}$

>>> 4ª equação:

→ Identificando os coeficientes:

$\large\begin {array}{l}\Large\boxed{\boxed{a=1, b= -2, c=-3}}\Huge\checkmark\end {array}$

→ discriminante (Δ):

$\large\begin {array}{l}\Delta=b^2-4\cdot a\cdot c\\\\\Delta=(-2)^2-4\cdot 1\cdot (-3)\\\\\Delta=4+12\\\\\Large\boxed{\boxed{\Delta=16}}\Huge\checkmark\end {array}$

→ Raízes:

$\large\begin {array}{l}x=\dfrac{-b\pm\sqrt{\Delta} }{2\cdot a}=\dfrac{-(-2)\pm\sqrt{16} }{2\cdot 1}=\dfrac{2\pm4 }{2}=\end {array}$

$\boxed{\large\begin {array}{l}x'=\dfrac{2+4}{2}\\\\x'=\dfrac{6}{2}\\\\\Large\boxed{\boxed{x'=\;3\;}}\Huge\checkmark\;\end {array}}\quad\quad \quad \boxed{\large\begin {array}{l}x"=\dfrac{2-4}{2}\\\\x"=\dfrac{-2}{2}\\\\\Large\boxed{\boxed{x"=-1}}\Huge\checkmark\end {array}}$

Resposta:

Portanto, a tabela preenchida ficará:

$\large\begin{array}{|l|c|c|c|c|l|} \cline{1-6}\text{Equa\c{c}\~ao}&"a"&"b"&"c"&\Delta&Ra\'izes\\ [3]\cline{1-6}f(x)=x^2+2x&1&2&0&4&x'=0\;e\;x"=-2\\ [3]\cline{1-6}f(x)=x^2-13x+40&1&-13&40&9&x'=8\;e\;x"=5\\ [3]\cline{1-6}f(x)=2x^2-7x+3&2&-7&3&25&x'=3\;e\;x"=0,5\\ [3]\cline{1-6}f(x)=x^2-2x-3&1&-2&-3&16&x'=3\;e\;x"=-1\\ [3]\cline{1-6} \end{array}$

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Bons estudos!