Question

me ajudem pfv :(
preciso mt

Answer 1

Resposta:

Explicação passo-a-passo:

Temos que pela fórmula de Baskhara, o zero de uma equação pode ser dado pela fórmula $x=\dfrac{-b\±\sqrt{\Delta}}{2\cdot{a}}$  Onde $\Delta=b^2-4\cdot{a}\cdot{c}$...

A)

$x=\dfrac{-4\±\sqrt{36}}{2}=\dfrac{-4\±6}{2}\\\\x_{1}=\dfrac{-4-6}{2}=\dfrac{-10}{2}=-5\\x_{2}=\dfrac{-4+6}{2}=\dfrac{2}{2}=1$

B)

$x=\dfrac{3\±{\sqrt{121}}}{2\cdot1}=\dfrac{3\±{11}}{2} \\,\\\Longrightarrow{x_{1}}=\dfrac{3-11}{2}=\dfrac{-8}{2}=-4\\\\\Longrightarrow{x_{2}}=\dfrac{3+11}{2}=\dfrac{14}{2}=7$

C)

$x=\dfrac{2\±\sqrt{64}}{2}=\dfrac{2\±8}{2}\\\\\Longrightarrow{x_{1}}=\dfrac{2+8}{2}=\dfrac{10}{2}=5\\\\\Longrightarrow{x_{2}}=\dfrac{2-8}{2}=\dfrac{-6}{2}=-3$

D)

$x=\dfrac{-3\±\sqrt{49}}{2}=\dfrac{-3\±7}{2}\\\\\Longrightarrow{x_{1}}=\dfrac{-3-7}{2}=\dfrac{-10}{2}=5\\\\\Longrightarrow{x_{2}}=\dfrac{-3+7}{2}=\dfrac{4}{2}=2$

E)

$x=\dfrac{-6\±\sqrt{64}}{2}=\dfrac{-6\±8}{2}\\\\\Longrightarrow{x_{1}}=\dfrac{-6-8}{2}=\dfrac{-14}{2}=-7\\\\\Longrightarrow{x_{2}}=\dfrac{-6+8}{2}=\dfrac{2}{2}=1$

F)

$x=\dfrac{-2\±\sqrt{16}}{2}=\dfrac{-2\±4}{2}\\\\\Longrightarrow{x_{1}}=\dfrac{-2-4}{2}=\dfrac{-6}{2}=-3\\\\\Longrightarrow{x_{2}}=\dfrac{-2+4}{2}=\dfrac{2}{2}=1$

G)

$x=\dfrac{10\±\sqrt{16}}{2}=\dfrac{10\±4}{2}=\dfrac{10}{2}\±\dfrac{4}{2}=5\±2\\\\\Longrightarrow{x_{1}}=5+2=7\\\\\Longrightarrow{x_{2}}=5-2=3$