Equações

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:
$$\begin{array}{*{20}{c}}
{A \times B = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}
{A = 0}& \vee &{B = 0}
\end{array}}
\end{array}$$

­

1. Resolvendo a equação, vem:
   $$\begin{array}{*{20}{l}}
   {3{x^2} – 7 = 0}& \Leftrightarrow &{{x^2} = \frac{7}{3}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = – \sqrt {\frac{7}{3}} }& \vee &{x = \sqrt {\frac{7}{3}} }
   \end{array}}
   \end{array}$$
2. Resolvendo a equação, vem:
   $$\begin{array}{*{20}{l}}
   {2\left( {{x^2} + x} \right) = x}& \Leftrightarrow &{2{x^2} + 2x – x = 0} \\
   {}& \Leftrightarrow &{x\left( {2x + 1} \right) = 0} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = 0}& \vee &{2x + 1 = 0}
   \end{array}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = 0}& \vee &{x = – \frac{1}{2}}
   \end{array}}
   \end{array}$$
3. Resolvendo a equação, vem:
   $$\begin{array}{*{20}{l}}
   {\frac{{13}}{4}{x^2} = \frac{{13}}{5}}& \Leftrightarrow &{\frac{{{x^2}}}{4} = \frac{1}{5}} \\
   {}& \Leftrightarrow &{{x^2} = \frac{4}{5}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = – \sqrt {\frac{4}{5}} }& \vee &{x = \sqrt {\frac{4}{5}} }
   \end{array}}
   \end{array}$$
4. Resolvendo a equação, vem:
   $$\begin{array}{*{20}{l}}
   {2{x^2} + 3 = 0}& \Leftrightarrow &{\overbrace {{x^2} = – \frac{3}{2}}^{{\text{Eq}}{\text{. impossivel}}}} \\
   {}& \Leftrightarrow &{x \in \emptyset }
   \end{array}$$
5. Resolvendo a equação, vem:
   $$\begin{array}{*{20}{l}}
   {\frac{4}{{\mathop 7\limits_{(1)} }}\left( {x – 2} \right)(x + 2) + \mathop x\limits_{(7)} = \frac{{9 + 7x}}{{\mathop 7\limits_{(1)} }}}& \Leftrightarrow &{4\left( {x – 2} \right)(x + 2) + 7x = 9 + 7x} \\
   {}& \Leftrightarrow &{4\left( {{x^2} – 4} \right) + 7x = 9 + 7x} \\
   {}& \Leftrightarrow &{4{x^2} – 16 = 9} \\
   {}& \Leftrightarrow &{{x^2} = \frac{{25}}{4}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = – \sqrt {\frac{{25}}{4}} }& \vee &{x = \sqrt {\frac{{25}}{4}} }
   \end{array}} \\
   {}& \Leftrightarrow &{\begin{array}{*{20}{l}}
   {x = – \frac{5}{2}}& \vee &{x = \frac{5}{2}}
   \end{array}}
   \end{array}$$
6. Resolvendo a equação, vem:
   \[\begin{array}{*{20}{l}}{x = 3{x^2}}& \Leftrightarrow &{3{x^2} – x = 0}\\{}& \Leftrightarrow &{x\left( {3x – 1} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{3x – 1 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = \frac{1}{3}}\end{array}}\end{array}\]
7. Resolvendo a equação, vem:
   \[\begin{array}{*{20}{l}}{4{x^2} – \frac{1}{4}x = 0}& \Leftrightarrow &{x\left( {4x – \frac{1}{4}} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{4x – \frac{1}{4} = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = \frac{1}{{16}}}\end{array}}\end{array}\]
8. Resolvendo a equação, vem:
   \[\begin{array}{*{20}{l}}{ – 2{x^2} – 5x = 0}& \Leftrightarrow &{ – x\left( {2x + 5} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{ – x = 0}& \vee &{2x + 5 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = – \frac{5}{2}}\end{array}}\end{array}\]
9. Resolvendo a equação, vem:
   \[\begin{array}{*{20}{l}}{\frac{1}{2}x – \frac{1}{5}{x^2} = 0}& \Leftrightarrow &{x\left( {\frac{1}{2} – \frac{1}{5}x} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{\frac{1}{2} – \frac{1}{5}x = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{5 – 2x = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = \frac{5}{2}}\end{array}}\end{array}\]
10. Resolvendo a equação, vem:
    \[\begin{array}{*{20}{l}}{\frac{4}{5}{{\left( {x – 2} \right)}^2} + x = \frac{{16 – 3x}}{5}}& \Leftrightarrow &{4{{\left( {x – 2} \right)}^2} + 5x = 16 – 3x}\\{}& \Leftrightarrow &{4\left( {{x^2} – 4x + 4} \right) + 5x = 16 – 3x}\\{}& \Leftrightarrow &{4{x^2} – 8x = 0}\\{}& \Leftrightarrow &{4x\left( {x – 2} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{4x = 0}& \vee &{x – 2 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = 2}\end{array}}\end{array}\]
11. Resolvendo a equação, vem:
    \[\begin{array}{*{20}{l}}{{{\left( {x – 1} \right)}^2} = 3{x^2} + 1}& \Leftrightarrow &{{x^2} – 2x + 1 = 3{x^2} + 1}\\{}& \Leftrightarrow &{2{x^2} + 2x = 0}\\{}& \Leftrightarrow &{2x\left( {x + 1} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{2x = 0}& \vee &{x + 1 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = – 1}\end{array}}\end{array}\]
12. Resolvendo a equação, vem:
    \[\begin{array}{*{20}{l}}{{{\left( {x – 3} \right)}^2} = \left( {x – 3} \right)\left( {x + 3} \right)}& \Leftrightarrow &{{{\left( {x – 3} \right)}^2} – \left( {x – 3} \right)\left( {x + 3} \right) = 0}\\{}& \Leftrightarrow &{\left( {x – 3} \right)\left[ {\left( {x – 3} \right) – \left( {x + 3} \right)} \right] = 0}\\{}& \Leftrightarrow &{\left( {x – 3} \right)\left( { – 6} \right) = 0}\\{}& \Leftrightarrow &{x = 3}\end{array}\]