Resolve as seguintes equações, utilizando o completamento do quadrado
Equações do 2.º grau: Matematicamente Falando 9 - Parte 2 Pág. 88 Ex. 1
Enunciado
Resolve as seguintes equações, utilizando o completamento do quadrado.
- \({x^2} – 4x + 4 = 1\)
- \(x\left( {x – 2} \right) = 6 – 3x\)
- \(4{x^2} – 13x + 3 = 0\)
Resolução
- Ora,
\[\begin{array}{*{20}{l}}{{x^2} – 4x + 4 = 1}& \Leftrightarrow &{{{\left( {x – 2} \right)}^2} = 1}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x – 2 = – 1}& \vee &{x – 2 = 1}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 1}& \vee &{x = 3}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ {1,\;3} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{x\left( {x – 2} \right) = 6 – 3x}& \Leftrightarrow &{{x^2} + x – 6 = 0}\\{}& \Leftrightarrow &{{{\left( {x + \frac{1}{2}} \right)}^2} – \frac{1}{4} – 6 = 0}\\{}& \Leftrightarrow &{{{\left( {x + \frac{1}{2}} \right)}^2} = \frac{{25}}{4}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x + \frac{1}{2} = – \frac{5}{2}}& \vee &{x + \frac{1}{2} = \frac{5}{2}}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 3}& \vee &{x = 2}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ { – 3,\;2} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{4{x^2} – 13x + 3 = 0}& \Leftrightarrow &{{{\left( {2x – \frac{{13}}{4}} \right)}^2} – \frac{{169}}{{16}} + 3 = 0}\\{}& \Leftrightarrow &{{{\left( {2x – \frac{{13}}{4}} \right)}^2} = \frac{{121}}{{16}}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{2x – \frac{{13}}{4} = – \frac{{11}}{4}}& \vee &{2x – \frac{{13}}{4} = \frac{{11}}{4}}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = \frac{1}{4}}& \vee &{x = 3}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ {\frac{1}{4},\;3} \right\}}\end{array}\]
