Qual é o conjunto-solução?
Monómios e polinómios: Matematicamente Falando 8 - Pág. 145 Ex. 3
Qual é o conjunto-solução de cada uma das seguintes equações?
| a) | \({x^2} = 9\) |
| b) | \({x^2} – 4 = 0\) |
| c) | \(4{x^2} – 25 = 0\) |
| d) | \(9{x^2} = 16\) |
| e) | \({x^2} + 1 = 0\) |
| f) | \( – 4{x^2} = – 100\) |
| g) | \(2{x^2} + 13 = 0\) |
| h) | \(64 = {x^2}\) |
O conjunto-solução de cada uma das seguintes equações é apresentado seguidamente.
| a) | \({x^2} = 9\) | \[\begin{array}{*{20}{l}}{{x^2} = 9}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \sqrt 9 }& \vee &{x = + \sqrt 9 }\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 3}& \vee &{x = 3}\end{array}}\end{array}\] | \[S = \left\{ { – 3,\;3} \right\}\] |
| b) | \({x^2} – 4 = 0\) | \[\begin{array}{*{20}{l}}{{x^2} – 4 = 0}& \Leftrightarrow &{{x^2} = 4}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \sqrt 4 }& \vee &{x = + \sqrt 4 }\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 2}& \vee &{x = 2}\end{array}}\end{array}\] | \[S = \left\{ { – 2,\;2} \right\}\] |
| c) | \(4{x^2} – 25 = 0\) | \[\begin{array}{*{20}{l}}{4{x^2} – 25 = 0}& \Leftrightarrow &{4{x^2} = 25}\\{}& \Leftrightarrow &{{x^2} = \frac{{25}}{4}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \sqrt {\frac{{25}}{4}} }& \vee &{x = + \sqrt {\frac{{25}}{4}} }\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \frac{5}{2}}& \vee &{x = \frac{5}{2}}\end{array}}\end{array}\] | \[S = \left\{ { – \frac{5}{2},\;\frac{5}{2}} \right\}\] |
| d) | \(9{x^2} = 16\) | \[\begin{array}{*{20}{l}}{9{x^2} = 16}& \Leftrightarrow &{{x^2} = \frac{{16}}{9}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \sqrt {\frac{{16}}{9}} }& \vee &{x = + \sqrt {\frac{{16}}{9}} }\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \frac{4}{3}}& \vee &{x = \frac{4}{3}}\end{array}}\end{array}\] | \[S = \left\{ { – \frac{4}{3},\;\frac{4}{3}} \right\}\] |
| e) | \({x^2} + 1 = 0\) | \[\begin{array}{*{20}{l}}{{x^2} + 1 = 0}& \Leftrightarrow &{{x^2} = – 1}\\{}& \Leftrightarrow &{x \in \emptyset }\end{array}\] | \[S = \left\{ {} \right\}\] |
| f) | \( – 4{x^2} = – 100\) | \[\begin{array}{*{20}{l}}{ – 4{x^2} = – 100}& \Leftrightarrow &{{x^2} = 25}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 5}& \vee &{x = 5}\end{array}}\end{array}\] | \[S = \left\{ { – 5,\;5} \right\}\] |
| g) | \(2{x^2} + 13 = 0\) | \[\begin{array}{*{20}{l}}{2{x^2} + 13 = 0}& \Leftrightarrow &{{x^2} = – \frac{{13}}{2}}\\{}& \Leftrightarrow &{x \in \emptyset }\end{array}\] | \[S = \left\{ {} \right\}\] |
| h) | \(64 = {x^2}\) | \[\begin{array}{*{20}{l}}{64 = {x^2}}& \Leftrightarrow &{{x^2} = 64}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 8}& \vee &{x = 8}\end{array}}\end{array}\] | \[S = \left\{ { – 8,\;8} \right\}\] |
Alternativa para a resolução da equação
Vamos resolver seguidamente algumas das equações acima, utilizando o caso notável \({a^2} – {b^2} = \left( {a + b} \right)\left( {a – b} \right)\) e a Lei do Anulamento do Produto (\(\begin{array}{*{20}{l}}{A \times B = 0}& \Leftrightarrow &{\begin{array}{*{20}{c}}{A = 0}& \vee &{B = 0}\end{array}}\end{array}\)).
\[\begin{array}{*{20}{l}}{{x^2} = 9}& \Leftrightarrow &{{x^2} – 9 = 0}\\{}& \Leftrightarrow &{\left( {x + 3} \right)\left( {x – 3} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x + 3 = 0}& \vee &{x – 3 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 3}& \vee &{x = 3}\end{array}}\end{array}\]
\[\begin{array}{*{20}{l}}{{x^2} – 4 = 0}& \Leftrightarrow &{\left( {x + 2} \right)\left( {x – 2} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x + 2 = 0}& \vee &{x – 2 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 2}& \vee &{x = 2}\end{array}}\end{array}\]
\[\begin{array}{*{20}{l}}{4{x^2} – 25 = 0}& \Leftrightarrow &{\left( {2x + 5} \right)\left( {2x – 5} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{2x + 5 = 0}& \vee &{2x – 5 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \frac{5}{2}}& \vee &{x = \frac{5}{2}}\end{array}}\end{array}\]
\[\begin{array}{*{20}{l}}{9{x^2} = 16}& \Leftrightarrow &{9{x^2} – 16 = 0}\\{}& \Leftrightarrow &{\left( {3x + 4} \right)\left( {3x – 4} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{3x + 4 = 0}& \vee &{3x – 4 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \frac{4}{3}}& \vee &{x = \frac{4}{3}}\end{array}}\end{array}\]
Há vantagem em utilizar esta alternativa?
