Determina o conjunto-solução das equações
Monómios e polinómios: Matematicamente Falando 8 - Pág. 148 Ex. 18
Enunciado
Determina o conjunto-solução das seguintes equações.
| a) | \(x\left( {x – 1} \right) = 0\) |
| b) | \(\left( {a – 1} \right)\left( {a + 1} \right) = 0\) |
| c) | \({x^2} – 2x = 0\) |
| d) | \(2{x^2} = 32\) |
| e) | \({c^2} – 0,25 = 0\) |
| f) | \({x^2} = 0,01\) |
| g) | \({y^3} – 4y = 0\) |
| h) | \({x^2} – 256 = 0\) |
Resolução
O conjunto-solução das equações está apresentado abaixo.
| a) | \[\begin{array}{*{20}{l}}{x\left( {x – 1} \right) = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x – 1 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = 1}\end{array}}\end{array}\] | \[S = \left\{ {0,\;1} \right\}\] |
| b) | \[\begin{array}{*{20}{l}}{\left( {a – 1} \right)\left( {a + 1} \right) = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}{a – 1 = 0}& \vee &{a + 1 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{a = 1}& \vee &{a = – 1}\end{array}}\end{array}\] | \[S = \left\{ { – 1,\;1} \right\}\] |
| c) | \[\begin{array}{*{20}{l}}{{x^2} – 2x = 0}& \Leftrightarrow &{x\left( {x – 2} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x – 2 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = 2}\end{array}}\end{array}\] | \[S = \left\{ {0,\;2} \right\}\] |
| d) | \[\begin{array}{*{20}{l}}{2{x^2} = 32}& \Leftrightarrow &{{x^2} = 16}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 4}& \vee &{x = 4}\end{array}}\end{array}\] | \[S = \left\{ { – 4,\;4} \right\}\] |
| e) | \[\begin{array}{*{20}{l}}{{c^2} – 0,25 = 0}& \Leftrightarrow &{{c^2} = 0,25}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 0,5}& \vee &{x = 0,5}\end{array}}\end{array}\] | \[S = \left\{ { – 0,5;\;0,5} \right\}\] |
| f) | \[\begin{array}{*{20}{l}}{{x^2} = 0,01}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 0,1}& \vee &{x = 0,1}\end{array}}\end{array}\] | \[S = \left\{ { – 0,1;\;0,1} \right\}\] |
| g) | \[\begin{array}{*{20}{l}}{{y^3} – 4y = 0}& \Leftrightarrow &{y\left( {{y^2} – 4} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{y = 0}& \vee &{{y^2} – 4 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{y = 0}& \vee &{{y^2} = 4}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{l}}{y = 0}& \vee &{y = – 2}& \vee &{y = 2}\end{array}}\end{array}\] | \[S = \left\{ { – 2,\;0,\;2} \right\}\] |
| h) | \[\begin{array}{*{20}{l}}{{x^2} – 256 = 0}& \Leftrightarrow &{{x^2} = 256}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \sqrt {256} }& \vee &{x = \sqrt {256} }\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 16}& \vee &{x = 16}\end{array}}\end{array}\] | \[S = \left\{ { – 16,\;16} \right\}\] |
