Utilizando o completamento do quadrado
Equações do 2.º grau: Matematicamente Falando 9 - Parte 2 Pág. 82 Ex. 4
Enunciado
Resolve as equações do 2.º grau, utilizando o completamento do quadrado.
- \({{x^2} – 6x + 5 = 0}\)
- \({{x^2} + 5x + 1 = 0}\)
Resolução
- Ora,
\[\begin{array}{*{20}{l}}{{x^2} – 6x + 5 = 0}& \Leftrightarrow &{{{\left( {x – 3} \right)}^2} – 9 + 5 = 0}\\{}& \Leftrightarrow &{{{\left( {x – 3} \right)}^2} = 4}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x – 3 = – 2}& \vee &{x – 3 = 2}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 1}& \vee &{x = 5}\end{array}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{{x^2} + 5x + 1 = 0}& \Leftrightarrow &{{{\left( {x + \frac{5}{2}} \right)}^2} – \frac{{25}}{4} + 1 = 0}\\{}& \Leftrightarrow &{{{\left( {x + \frac{5}{2}} \right)}^2} = \frac{{21}}{4}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x + \frac{5}{2} = – \frac{{\sqrt {21} }}{2}}& \vee &{x + \frac{5}{2} = \frac{{\sqrt {21} }}{2}}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = \frac{{ – 5 – \sqrt {21} }}{2}}& \vee &{x = \frac{{ – 5 + \sqrt {21} }}{2}}\end{array}}\end{array}\]
