Question

Aplicando a fórmula de bhaskara, resolva as seguintes equações do 2° grau.

Answer 1

Resposta:

h) x' = 2 e x'' = -2

i) x' = 1 e x'' = -2

Explicação passo-a-passo:

h)

$3x^2-12=0\\\\x=\dfrac{-b\pm\sqrt{b^2 - 4\ .\ a\ .\ c}}{2\ .\ a}\\\\x=\dfrac{-0\pm\sqrt{0^2 - 4\ .\ 3\ .\ -12}}{2\ .\ 3}=\dfrac{\pm\sqrt{+144}}{6} = \dfrac{\pm12}{6}\\\\x'=\dfrac{+12}{6}=2 \Rightarrow \boxed{\bf{x'=2}}\\\\x'=\dfrac{-12}{6}=-2 \Rightarrow \boxed{\bf{x''=-2}}$

i)

$x^2+x-2=0\\\\x=\dfrac{-b\pm\sqrt{b^2 - 4\ .\ a\ .\ c}}{2\ .\ a}\\\\x=\dfrac{-1\pm\sqrt{1^2 - 4\ .\ 1\ .\ -2}}{2\ .\ 1}=\dfrac{-1\pm\sqrt{+9}}{2} = \dfrac{-1\pm3}{2}\\\\x'=\dfrac{-1+3}{2}=\dfrac{2}{2}=1 \Rightarrow \boxed{\bf{x'=1}}\\\\x'=\dfrac{-1-3}{2}=\dfrac{-4}{2}=-2 \Rightarrow \boxed{\bf{x''=-2}}$

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Answer 2

Explicação passo-a-passo:

h)

$\sf 3x^2-12=0$

$\sf \Delta=0^2-4\cdot3\cdot(-12)$

$\sf \Delta=0+144$

$\sf \Delta=144$

$\sf x=\dfrac{-0\pm\sqrt{144}}{2\cdot3}=\dfrac{0\pm12}{6}$

$\sf x'=\dfrac{12}{6}~\Rightarrow~\red{x'=2}$

$\sf x'=\dfrac{-12}{6}~\Rightarrow~\red{x'=-2}$

O conjunto solução é $\sf S=\{-2,2\}$

i)

$\sf x^2+x-2=0$

$\sf \Delta=1^2-4\cdot1\cdot(-2)$

$\sf \Delta=1+8$

$\sf \Delta=9$

$\sf x=\dfrac{-1\pm\sqrt{9}}{2\cdot1}=\dfrac{-1\pm3}{2}$

$\sf x'=\dfrac{-1+3}{2}~\Rightarrow~x'=\dfrac{2}{2}~\Rightarrow~\red{x'=1}$

$\sf x"=\dfrac{-1-3}{2}~\Rightarrow~x"=\dfrac{-4}{2}~\Rightarrow~\red{x"=-2}$

O conjunto solução é $\sf S=\{-2,1\}$