Question

Qual é  o número cuja o dobro somado com seu qradrado é igual a 8

Answer 1

$[tex]2x+ x^{2} =8$$x^{2} +2x-8=0$/$delta= 2^{2}-4.1.-8 /delta=4+32/delta=36$[/tex]
$x= \frac{-(4)mais/menos \sqrt{36} }{2.1}$
$x= \frac{-4mais/menos6}{2}$
$x^{1} = \frac{4+6}{2}$$[tex] x^{1}=5$[/tex]
$x^{2} = \frac{4-6}{2}$ $x^{2} =-1$

Answer 2

$2x+x^2=8\Longrightarrow x^2+2x-8=0\\\\ \Delta=b^2-4\cdot a\cdot c\\ \Delta=2^2-4\cdot1\cdot(-8)\\ \Delta=4+32\\ \Delta=36\\\\ x=\dfrac{-b\pm\sqrt{\Delta}}{2a}=\dfrac{-2\pm\sqrt{36}}{2\cdot1}=\dfrac{-2\pm6}{2}=-1\pm3\Longrightarrow\begin{cases}x_1=2\\x_2=-4\end{cases}\\\\ S=\{-4,2\}$