Question

Como fazer x²+5x+6=0

Answer 1

x² = a
5x = b
6 = c
Delta = b² - 4.a.c
Delta = 5² - 4.1.6
             25 - 24 = 1
x= -b +ou- raiz de delta / 2.a
x= -5 +ou- 1 / 2
x1: -5 +1= -4/2 = -2
x2: -5 -1= -6/2 = -3

Answer 2

$\textsf{Equa\c{c}\~ao quadr\'atica:}~~\mathsf{ax^2+bx+c=0}\\\\\\\textsf{A figura geom\'etrica que representa essa equa\c{c}\~ao \'e a par\'abola; e ao igu\underline{a}}\\\textsf{larmos a equa\c{c}\~ao a zero temos como objetivo descobrir suas ra\'izes, po\underline{n}}\\\textsf{tos onde a par\'abola intercepta/corta o eixo x.}\\\\\\\textsf{Para resolver a equa\c{c}\~ao podemos usar da seguinte f\'ormula:}\\\\\\\textsf{F\'ormula de Bhaskara:}~~\mathsf{ \dfrac{-b\pm \sqrt{\Delta} }{2a},~onde~\Delta=b^2-4ac}$


$\mathsf{A~equa\c{c}\~ao~que~voc\^e~prop\^os~\'e~a~seguinte:~x^2+5x+6=0}\\\\\therefore~~\mathsf{a=1,~~b=5,~~c=6}\\\\\\\mathsf{Calculando~\Delta:}\\\\\\\Delta=5^2-4\cdot 1\cdot 6\\\\\mathsf{\Delta=25-24}\\\\\mathsf{\Delta=1}\\\\\\\mathsf{Substituindo~\Delta~na~f\'ormula~de~Bhaskara:}\\\\\\\mathsf{\dfrac{-5\pm\sqrt{1}}{2\cdot 1}}\\\\\\\mathsf{\dfrac{-5\pm{1}}{2}}$


$\mathsf{\dfrac{-5\pm1}{2}}~\Rightarrow~\left\{\begin{matrix} \mathsf{\dfrac{-5+1}{2}=-\dfrac{4}{2}=-2}}\\\\\\\\\mathsf{\dfrac{-5-1}{2}= -\dfrac{6}{2}=-3} \end{matrix}\right.}\\\\\\\\\\\mathsf{\therefore~~~x}=\left\{\begin{matrix} \mathsf{-2}\\\\\mathsf{ou}\\\\\mathsf{-3} \end{matrix}\right.$



$\mathsf{Obs:~quando~\Delta~\'e~positivo~(\Delta\ \textgreater \ 0)~a~par\'abola~intercepta~o~eixo~x}\\\textsf{em dois pontos, ou seja, apresenta duas ra\'izes.}\\\\\textsf{Outra informa\c{c}\~ao importante \'e em rela\c{c}\~ao a concavidade da par\'abola;}\\\mathsf{quando~a\ \textgreater \ 0~a~concavidade~\'e~para~cima~(ex:~\cup),~j\'a~quando~a\ \textless \ 0}\\\mathsf{a~concavidade~\'e~voltada~para~baixo~(ex:\cap).}$


$\textsf{Imagem da par\'abola em anexo:}$