Resolve as seguintes equações em x
Equações do 2.º grau: Matematicamente Falando 9 - Parte 2 Pág. 88 Ex. 7
Enunciado
Resolve as seguintes equações em x.
- \({3{x^2} + 5x = 0}\)
- \({\sqrt 2 {x^2} + 11x = 0}\)
- \({{x^2} + 9 = 0}\)
- \({\left( {x – 4} \right)\left( {x + 1} \right) = 10 – 3x}\)
- \({{x^2} – 10x + 24 = 0}\)
- \({{x^2} – 4x = – 3}\)
- \({\left( {x + 4} \right)\left( {x – 1} \right) = 5x – 20}\)
- \({5x + {{\left( {x + 2} \right)}^2} = 3x\left( {x + 2} \right) + x}\)
- \({\left( {x + 2} \right)\left( {x – 2} \right) – {{\left( {x – 1} \right)}^2} = {x^2} – 8}\)
- \({\frac{{{x^2} – 1}}{4} = \frac{{x – 1}}{3}}\)
- \(\frac{{{x^2}}}{2} – 1 = \frac{x}{3} + 15\)
- \({4,8{x^2} – 8,4x + 2,4 = 0}\)
- \({\frac{{x – 1}}{2} – \frac{{x\left( {3 – x} \right)}}{3} = x + \frac{1}{3}}\)
Resolução
- Ora,
\[\begin{array}{*{20}{l}}{3{x^2} + 5x = 0}& \Leftrightarrow &{x\left( {3x + 5} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{3x + 5 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = – \frac{5}{3}}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ { – \frac{5}{3},\;0} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{\sqrt 2 {x^2} + 11x = 0}& \Leftrightarrow &{x\left( {\sqrt 2 x + 11} \right) = 0}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{\sqrt 2 x + 11 = 0}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = – \frac{{11}}{{\sqrt 2 }}}\end{array}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 0}& \vee &{x = – \frac{{11\sqrt 2 }}{2}}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ { – \frac{{11\sqrt 2 }}{2},\;0} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{{x^2} + 9 = 0}& \Leftrightarrow &{{x^2} = – 9}\\{}& \Leftrightarrow &{x \in \emptyset }\\{}&{}&{}\\{}&{}&{S = \left\{ {} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{\left( {x – 4} \right)\left( {x + 1} \right) = 10 – 3x}& \Leftrightarrow &{{x^2} + x – 4x – 4 – 10 + 3x = 0}\\{}& \Leftrightarrow &{{x^2} – 14 = 0}\\{}& \Leftrightarrow &{{x^2} = 14}\\{}& \Leftrightarrow &{x = \mp \sqrt {14} }\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \sqrt {14} }& \vee &{x = \sqrt {14} }\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ { – \sqrt {14} ,\;\sqrt {14} } \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{{x^2} – 10x + 24 = 0}& \Leftrightarrow &{x = \frac{{10 \mp \sqrt {100 – 96} }}{2}}\\{}& \Leftrightarrow &{x = \frac{{10 \mp 2}}{2}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 4}& \vee &{x = 6}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ {4,\;6} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{{x^2} – 4x = – 3}& \Leftrightarrow &{{x^2} – 4x + 3 = 0}\\{}& \Leftrightarrow &{x = \frac{{4 \mp \sqrt {16 – 12} }}{2}}\\{}& \Leftrightarrow &{x = \frac{{4 \mp 2}}{2}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = 1}& \vee &{x = 3}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ {1,\;3} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{\left( {x + 4} \right)\left( {x – 1} \right) = 5x – 20}& \Leftrightarrow &{{x^2} – x + 4x – 4 – 5x + 20 = 0}\\{}& \Leftrightarrow &{\underbrace {{x^2} – 2x + 16 = 0}_{\Delta = 4 – 64 < 0}}\\{}& \Leftrightarrow &{x \in \emptyset }\\{}&{}&{}\\{}&{}&{S = \left\{ {} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{5x + {{\left( {x + 2} \right)}^2} = 3x\left( {x + 2} \right) + x}& \Leftrightarrow &{5x + {x^2} + 4x + 4 – 3{x^2} – 6x – x = 0}\\{}& \Leftrightarrow &{ – 2{x^2} + 2x + 4 = 0}\\{}& \Leftrightarrow &{{x^2} – x – 2 = 0}\\{}& \Leftrightarrow &{x = \frac{{1 \mp \sqrt {1 + 8} }}{2}}\\{}& \Leftrightarrow &{x = \frac{{1 \mp 3}}{2}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 1}& \vee &{x = 2}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ { – 1,\;2} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{\left( {x + 2} \right)\left( {x – 2} \right) – {{\left( {x – 1} \right)}^2} = {x^2} – 8}& \Leftrightarrow &{{x^2} – 4 – {x^2} + 2x – 1 – {x^2} + 8 = 0}\\{}& \Leftrightarrow &{ – {x^2} + 2x + 3 = 0}\\{}& \Leftrightarrow &{x = \frac{{ – 2 \mp \sqrt {4 + 12} }}{{ – 2}}}\\{}& \Leftrightarrow &{x = \frac{{ – 2 \mp 4}}{{ – 2}}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – 1}& \vee &{x = 3}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ { – 1,\;3} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{\frac{{{x^2} – 1}}{4} = \frac{{x – 1}}{3}}& \Leftrightarrow &{3{x^2} – 3 = 4x – 4}\\{}& \Leftrightarrow &{3{x^2} – 4x + 1 = 0}\\{}& \Leftrightarrow &{x = \frac{{4 \mp \sqrt {16 – 12} }}{6}}\\{}& \Leftrightarrow &{x = \frac{{4 \mp 2}}{6}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = \frac{1}{3}}& \vee &{x = 1}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ {\frac{1}{3},\;1} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{\frac{{{x^2}}}{{\mathop 2\limits_{\left( 3 \right)} }} – \mathop 1\limits_{\left( 6 \right)} = \frac{x}{{\mathop 3\limits_{\left( 2 \right)} }} + \mathop {15}\limits_{\left( 6 \right)} }& \Leftrightarrow &{3{x^2} – 6 = 2x + 90}\\{}& \Leftrightarrow &{3{x^2} – 2x – 96 = 0}\\{}& \Leftrightarrow &{x = \frac{{2 \mp \sqrt {4 + 1152} }}{6}}\\{}& \Leftrightarrow &{x = \frac{{2 \mp 34}}{6}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \frac{{16}}{3}}& \vee &{x = 6}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ { – \frac{{16}}{3},\;6} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{4,8{x^2} – 8,4x + 2,4 = 0}& \Leftrightarrow &{4{x^2} – 7x + 2 = 0}\\{}& \Leftrightarrow &{x = \frac{{7 \mp \sqrt {49 – 32} }}{8}}\\{}& \Leftrightarrow &{x = \frac{{7 \mp \sqrt {17} }}{8}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = \frac{{7 – \sqrt {17} }}{8}}& \vee &{x = \frac{{7 + \sqrt {17} }}{8}}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ {\frac{{7 – \sqrt {17} }}{8},\;\frac{{7 + \sqrt {17} }}{8}} \right\}}\end{array}\] - Ora,
\[\begin{array}{*{20}{l}}{\frac{{x – 1}}{2} – \frac{{x\left( {3 – x} \right)}}{3} = x + \frac{1}{3}}& \Leftrightarrow &{3x – 3 – 6x + 2{x^2} – 6x – 2 = 0}\\{}& \Leftrightarrow &{2{x^2} – 9x – 5 = 0}\\{}& \Leftrightarrow &{x = \frac{{9 \mp \sqrt {81 + 40} }}{4}}\\{}& \Leftrightarrow &{x = \frac{{9 \mp 11}}{4}}\\{}& \Leftrightarrow &{\begin{array}{*{20}{c}}{x = – \frac{1}{2}}& \vee &{x = 5}\end{array}}\\{}&{}&{}\\{}&{}&{S = \left\{ { – \frac{1}{2},\;5} \right\}}\end{array}\]
