Question

Aplique a fórmula de bháskara e encontre suas raízes:
${x}^{2} + 3x + 2 = 0$
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Answer 1

Resposta:

Explicação passo a passo:

$\displaystyle Aplicando~a~f\acute{o}rmula~de~Bhaskara~para~x^{2}+3x+2=0~~e~comparando~com~(a)x^{2}+(b)x+(c)=0,~determinamos~os~coeficientes:~\\a=1{;}~b=3~e~c=2\\\\C\acute{a}lculo~do~discriminante~(\Delta):&\\&~\Delta=(b)^{2}-4(a)(c)=(3)^{2}-4(1)(2)=9-(8)=1\\\\C\acute{a}lculo~das~raizes:&\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(3)-\sqrt{1}}{2(1)}=\frac{-3-1}{2}=\frac{-4}{2}=-2\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(3)+\sqrt{1}}{2(1)}=\frac{-3+1}{2}=\frac{-2}{2}=-1\\\\S=\{-2,~-1\}$

Answer 2

$\boxed{\begin{array}{lr}x {}^{2} + 3x + 2 = 0\\ \\\begin{cases}a = 1 \\ b = 3 \\ c = 2\end{cases} \\ \\∆ = b {}^{2} - 4.a.c \\ ∆ = 3 {}^{2} - 4.1.2 \\ ∆ = 9 - 8 \\ ∆ = 1 \\ \\ x = \frac{ - b± \sqrt{∆} }{2.a} \\ \\ x = \frac{ - 3± \sqrt{1} }{2.1} \\ \\ x = \frac{ - 3±1}{2} \\ \\ x_{1} = \frac{ - 3 + 1}{2} = \frac{ - 2}{2} = \boxed{ - 1} \\ \\ x_{2} = \frac{ - 3 - 1}{2} = \frac{ - 4}{2} = \boxed{ - 2} \\ \\ \boxed{s =[ - 1,-2] }\end{array}}$