Question

determine i cojunto-solução das seguintes equação do 2 grau no conjunto ir
${x}^{2} - 7x + 10 \\ \\ (x - 5) {}^{2} = 25 - 9x$
${x}^{2} - \frac{4x}{5} = \frac{1}{5}$
$\frac{x {}^{2} }{2} - \frac{x + 12}{ 3} = 2x$

Answer 1

$x^2-7x+10=0\\\\ \Delta=(-7)^2-4.1.10\\ \Delta=49-10=9\\\\ x=\frac{-(-7)\pm\sqrt{9}}{2}=\frac{7\pm3}{2}\\\\ x_1=\frac{7+3}{2}=\frac{10}{2}=5\\\\ x_2=\frac{7-3}{2}=\frac{4}{2}=2$


$(x-5)^2=25-9x\\\\ x^2-10x+25=25-9x\\ x^2-10x+25-25+9x=0\\ x^2-x=0\\ x(x-1)=0\\ x=0\\ x-1=0\\ x=1$

$x^2-\frac{4x}{5}=\frac{1}{5}\\\\ 5x^2-4x=1\\ 5x^2-4x-1=0\\ \Delta=(-4)^2-4.5.(-1)\\ \Delta=16+20=36\\\\ x=\frac{-(-4)\pm\sqrt{36}}{2.5}=\frac{4\pm6}{10}\\\\ x_1=\frac{4+6}{10}=\frac{10}{10}=1\\\\ x_2=\frac{4-6}{10}=\frac{-2}{10}=\frac{-1}{5}$


$\frac{x^2}{2}-\frac{x +12}{3}=2x\\\\ \frac{3x^2-(2).(x +12)=6.2x}{6}\\\\ 3x^2-(2).(x +12)=6.2x\\ 3x^2-2x -24=12x\\ 3x^2-14x -24=0\\ \Delta=(-14)^2-4.3.(-24)\\ \Delta=196+288=484\\\\ x=\frac{-(-14)\pm\sqrt{484}}{2.3}=\frac{14\pm22}{6}\\\\ x_1=\frac{14+22}{6}=\frac{36}{6}=6\\\\ x_2=\frac{14-22}{6}=\frac{-8}{6}=\frac{-4}{3}$