Question

Resolva o seguinte sistema {x+y=38 x+2=(y+2)²

Answer 1

$\left \{ {{x\,+\,y\,=\,38} \atop {x\,+\,2\,=\,(y+2)^{2}}} \right.$

$\left \{ {{x\,=\,38\,-\,y} \atop {x\,+\,2\,=\,(y+2)^{2}}} \right.$

$x+2=(y+2)^{2}$

$38 - y + 2 = (y+2)^{2}$

$38 - y + 2 = y^{2} + 4y + 4$

$y^{2} + 4y + y + 4 - 2 - 38 = 0$

$y^{2} + 5y - 36 = 0$

$Equa\c{c}\~ao\;de\;segundo\;grau\;igual\;a\;zero.\;F\'ormula\;de\;Bhaskara!$

$y = \frac{-b\,\pm \sqrt{b^{2}-4ac}}{2a}$

$y = \frac{-5\,\pm \sqrt{5^{2}-4.1.(-36)}}{2.1}$

$y = \frac{-5\,\pm \sqrt{25-(-144)}}{2}$

$y = \frac{-5\,\pm \sqrt{25+144}}{2}$

$y = \frac{-5\,\pm \sqrt{169}}{2}$

$y = \frac{-5\,\pm 13}{2}$

$y' = \frac{-5\,+\, 13}{2}$

$y' = \frac{8}{2}$

$y' = 4$

$y'' = \frac{-5\,-\, 13}{2}$

$y'' = \frac{-18}{2}$

$y'' = -9$

$\fbox{\fbox{S = 4\,;\,-9}}$

$Ent\~ao,\;usemos\;os\;valores\;encontrados\;para\;\,'\!y'\;na\;equa\c{c}\~ao:$

$Se\; y = 4,\;temos\;que$

$x=38-y$

$x=38-4$

$\fbox{\fbox{x = 34}}$

$Se\; y = -9,\;temos\;que$

$x=38-(\!-9)$

$x=38+9$

$\fbox{\fbox{x = 47}}$