Resolve as equações
Monómios e polinómios: Matematicamente Falando 8 - Pág. 144 Ex. 2
Enunciado
Resolve as equações, utilizando a lei do anulamento do produto.
| a) | \(x\left( {x + 2} \right) = 0\) |
| b) | \(\left( {2x + 1} \right)\left( {x – \frac{1}{3}} \right) = 0\) |
| c) | \({x^2} + 3x = 0\) |
| d) | \(3{z^2} – 12z = 0\) |
| e) | \(\left( {x – 3} \right)\left( {2 + 7x} \right) = 0\) |
| f) | \(x\left( {x + 1} \right) + 2\left( {x + 1} \right) = 0\) |
| g) | \( – x\left( {x + 4} \right) = 0\) |
| h) | \(\left( {x + 4} \right)x – 3\left( {x + 4} \right) = 0\) |
| i) | \(3{x^2} – 12 = 0\) |
| j) | \(16x + 2{x^2} = 0\) |
| k) | \(2{m^2} + 5m = 0\) |
Resolução
As equações estão resolvidas por utilização da lei do anulamento do produto.
| a) | \(\begin{array}{*{20}{l}}{x\left( {x + 2} \right) = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x + 2 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = – 2}\end{array}}\end{array}\) |
| b) | \(\begin{array}{*{20}{l}}{\left( {2x + 1} \right)\left( {x – \frac{1}{3}} \right) = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}{2x + 1 = 0}& \vee &{x – \frac{1}{3} = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \frac{1}{2}}& \vee &{x = \frac{1}{3}}\end{array}}\end{array}\) |
| c) | \(\begin{array}{*{20}{l}}{{x^2} + 3x = 0}& \Leftrightarrow &{x\left( {x + 3} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x + 3 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = – 3}\end{array}}\end{array}\) |
| d) | \(\begin{array}{*{20}{l}}{3{z^2} – 12z = 0}& \Leftrightarrow &{3z\left( {z – 4} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{3z = 0}& \vee &{z – 4 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{z = 0}& \vee &{z = 4}\end{array}}\end{array}\) |
| e) | \(\begin{array}{*{20}{l}}{\left( {x – 3} \right)\left( {2 + 7x} \right) = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x – 3 = 0}& \vee &{2 + 7x = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 3}& \vee &{x = – \frac{2}{7}}\end{array}}\end{array}\) |
| f) | \(\begin{array}{*{20}{l}}{x\left( {x + 1} \right) + 2\left( {x + 1} \right) = 0}& \Leftrightarrow &{\left( {x + 1} \right)\left( {x + 2} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x + 1 = 0}& \vee &{x + 2 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 1}& \vee &{x = – 2}\end{array}}\end{array}\) |
| g) | \(\begin{array}{*{20}{l}}{ – x\left( {x + 4} \right) = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}{ – x = 0}& \vee &{x + 4 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = – 4}\end{array}}\end{array}\) |
| h) | \(\begin{array}{*{20}{l}}{\left( {x + 4} \right)x – 3\left( {x + 4} \right) = 0}& \Leftrightarrow &{\left( {x + 4} \right)\left( {x – 3} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x + 4 = 0}& \vee &{x – 3 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 4}& \vee &{x = 3}\end{array}}\end{array}\) |
| i) | \(\begin{array}{*{20}{l}}{3{x^2} – 12 = 0}& \Leftrightarrow &{3\left( {{x^2} – 4} \right) = 0}\\{}& \Leftrightarrow &{3\left( {x + 2} \right)\left( {x – 2} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x + 2 = 0}& \vee &{x – 2 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 2}& \vee &{x = 2}\end{array}}\end{array}\) |
| j) | \(\begin{array}{*{20}{l}}{16x + 2{x^2} = 0}& \Leftrightarrow &{2x\left( {8 + x} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{2x = 0}& \vee &{8 + x = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = – 8}\end{array}}\end{array}\) |
| k) | \(\begin{array}{*{20}{l}}{2{m^2} + 5m = 0}& \Leftrightarrow &{m\left( {2m + 5} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{m = 0}& \vee &{2m + 5 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{m = 0}& \vee &{m = – \frac{5}{2}}\end{array}}\end{array}\) |
![Um triângulo [ABC]](https://www.acasinhadamatematica.pt/wp-content/uploads/2017/10/9V1Pag040-1_520x245.png)
