Resolve as seguintes equações do 2.º grau, utilizando o completamento do quadrado
Equações do 2.º grau: Matematicamente Falando 9 - Parte 2 Pág. 86 Ex. 7
Enunciado
Resolve as seguintes equações do 2.º grau, utilizando o completamento do quadrado.
- \({x^2} + 2x – 3 = 0\)
- \({x^2} – 13x + 42 = 0\)
- \( – {x^2} – 5x + 3 = 0\)
- \(3{x^2} + 5x = 2\)
Resolução
- Ora,
\[\begin{array}{*{20}{l}}{{x^2} + 2x – 3 = 0}& \Leftrightarrow &{{{\left( {x + 1} \right)}^2} – 1 – 3 = 0}\\{}& \Leftrightarrow &{{{\left( {x + 1} \right)}^2} = 4}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x + 1 = – 2}& \vee &{x + 1 = 2}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 3}& \vee &{x = 1}\end{array}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{{x^2} – 13x + 42 = 0}& \Leftrightarrow &{{{\left( {x – \frac{{13}}{2}} \right)}^2} – \frac{{169}}{4} + 42 = 0}\\{}& \Leftrightarrow &{{{\left( {x – \frac{{13}}{2}} \right)}^2} = \frac{1}{4}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x – \frac{{13}}{2} = – \frac{1}{2}}& \vee &{x – \frac{{13}}{2} = \frac{1}{2}}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 6}& \vee &{x = 7}\end{array}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{ – {x^2} – 5x + 3 = 0}& \Leftrightarrow &{{x^2} + 5x – 3 = 0}\\{}& \Leftrightarrow &{{{\left( {x + \frac{5}{2}} \right)}^2} – \frac{{25}}{4} – 3 = 0}\\{}& \Leftrightarrow &{{{\left( {x + \frac{5}{2}} \right)}^2} = \frac{{37}}{4}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x + \frac{5}{2} = – \frac{{\sqrt {37} }}{2}}& \vee &{x + \frac{5}{2} = \frac{{\sqrt {37} }}{2}}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = \frac{{ – 5 – \sqrt {37} }}{2}}& \vee &{x = \frac{{ – 5 + \sqrt {37} }}{2}}\end{array}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{3{x^2} + 5x = 2}& \Leftrightarrow &{{x^2} + \frac{5}{3}x – \frac{2}{3} = 0}\\{}& \Leftrightarrow &{{{\left( {x + \frac{5}{6}} \right)}^2} – \frac{{25}}{{36}} – \frac{2}{3} = 0}\\{}& \Leftrightarrow &{{{\left( {x + \frac{5}{6}} \right)}^2} = \frac{{49}}{{36}}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x + \frac{5}{6} = – \frac{7}{6}}& \vee &{x + \frac{5}{6} = \frac{7}{6}}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 2}& \vee &{x = \frac{1}{3}}\end{array}}\end{array}\]