Resolve as seguintes equações
Equações do 2.º grau: Matematicamente Falando 9 - Pág. 60 Ex. 4
Enunciado
Resolve as seguintes equações:
- $3{x^2} – 7 = 0$
- $2\left( {{x^2} + x} \right) = x$
- $\frac{{13}}{4}{x^2} = \frac{{13}}{5}$
- $2{x^2} + 3 = 0$
- $\frac{4}{7}\left( {x – 2} \right)(x + 2) + x = \frac{{9 + 7x}}{7}$
Resolução
Casos notáveis:
$${\left( {A + B} \right)^2} = {A^2} + 2AB + {B^2}$$$$\left( {A + B} \right)\left( {A – B} \right) = {A^2} – {B^2}$$
Lei do anulamento do produto:
$$\begin{array}{*{20}{c}}
{A \times B = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{A = 0}& \vee &{B = 0}
\end{array}}
\end{array}$$
- Resolvendo a equação, temos:
$$\begin{array}{*{20}{l}}
{3{x^2} – 7 = 0}& \Leftrightarrow &{{x^2} = \frac{7}{3}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{x = – \sqrt {\frac{7}{3}} }& \vee &{x = \sqrt {\frac{7}{3}} }
\end{array}}
\end{array}$$ - Resolvendo a equação, temos:
$$\begin{array}{*{20}{l}}
{2\left( {{x^2} + x} \right) = x}& \Leftrightarrow &{2{x^2} + 2x – x = 0} \\
{}& \Leftrightarrow &{x\left( {2x + 1} \right) = 0} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{x = 0}& \vee &{2x + 1 = 0}
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{x = 0}& \vee &{x = – \frac{1}{2}}
\end{array}}
\end{array}$$ - Resolvendo a equação, temos:
$$\begin{array}{*{20}{l}}
{\frac{{13}}{4}{x^2} = \frac{{13}}{5}}& \Leftrightarrow &{\frac{{{x^2}}}{4} = \frac{1}{5}} \\
{}& \Leftrightarrow &{{x^2} = \frac{4}{5}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{x = – \sqrt {\frac{4}{5}} }& \vee &{x = \sqrt {\frac{4}{5}} }
\end{array}}
\end{array}$$ - Resolvendo a equação, temos:
$$\begin{array}{*{20}{l}}
{2{x^2} + 3 = 0}& \Leftrightarrow &{\overbrace {{x^2} = – \frac{3}{2}}^{{\text{Eq}}{\text{. impossivel}}}} \\
{}& \Leftrightarrow &{x \in \emptyset }
\end{array}$$ - Resolvendo a equação, temos:
$$\begin{array}{*{20}{l}}
{\frac{4}{{\mathop 7\limits_{(1)} }}\left( {x – 2} \right)(x + 2) + \mathop x\limits_{(7)} = \frac{{9 + 7x}}{{\mathop 7\limits_{(1)} }}}& \Leftrightarrow &{4\left( {x – 2} \right)(x + 2) + 7x = 9 + 7x} \\
{}& \Leftrightarrow &{4\left( {{x^2} – 4} \right) + 7x = 9 + 7x} \\
{}& \Leftrightarrow &{4{x^2} – 16 = 9} \\
{}& \Leftrightarrow &{{x^2} = \frac{{25}}{4}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{x = – \sqrt {\frac{{25}}{4}} }& \vee &{x = \sqrt {\frac{{25}}{4}} }
\end{array}} \\
{}& \Leftrightarrow &{\begin{array}{*{20}{l}}
{x = – \frac{5}{2}}& \vee &{x = \frac{5}{2}}
\end{array}}
\end{array}$$
