Question

RESOLVA AS EQUAÇÕES DO 2º GRAU, USANDO BHÁSKARA
A) X2 –X – 2 =0
B) X2 + 2X + 1=0
C) X2 + 3X – 10 =0​

Answer 1

$\Box \ \ \boxed{\begin{array}{l}\sf a) \ x^2-x-2=0 \end{array}}\\\\\\\Box \ \ \boxed{\begin{array}{l}\sf a=1 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf b=-1 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf c=-2 \end{array}}\\\\\\\Box \ \ \boxed{\begin{array}{l}\sf \Delta\ =b^2-4 \ . \ a \ . \ c \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \Delta=(-1^2)-4 \ . \ 1 \ . \ -2 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \Delta=1+8 \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf \Delta =9 \ \checkmark \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf x=\frac{-b\pm\sqrt{\Delta}}{2.a} \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf x= \frac{1\pm\sqrt{9}}{2.1} \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf x=\frac{1\pm3}{2} \end{array}}\\\\\\\\\Box \ \ \boxed{\begin{array}{l}\sf x'=\frac{1+3}{2} \end{array}}\boxed{\begin{array}{l}\sf x'=\frac{4}{2} \end{array}}\boxed{\begin{array}{l}\sf x'=2 \ \checkmark \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf x''=\frac{1-3}{2} \end{array}}\boxed{\begin{array}{l}\sf x''=\frac{-2}{2} \end{array}}\boxed{\begin{array}{l}\sf x''=-1 \ \checkmark \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf S=\{ -1 \ , \ 2\} \end{array}}$

Resposta correta da alternativa a) ;

$\boxed{\begin{array}{l}\sf S=\{ -1 \ , \ 2\} \end{array}}$

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$\Box \ \ \boxed{\begin{array}{l}\sf b) \ x^2+2x+1=0 \end{array}}\\\\\\\\\Box \ \ \boxed{\begin{array}{l}\sf a=1 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf b=2 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf c=1 \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf \Delta=b^2-4\ . \ a\ .\ c \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \Delta=(2^2)-4\ .\ 1\ .\ 1 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \Delta=4-4 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \Delta=0 \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf x=\frac{-b\pm\sqrt{\Delta}}{2.a} \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf x=\frac{-2\pm\sqrt{0}}{2.1} \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf x=\frac{-2\pm 0}{2} \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf x'=\frac{-2\ +\ 0}{2} \end{array}}\boxed{\begin{array}{l}\sf x'=\frac{-2}{2} \end{array}}\boxed{\begin{array}{l}\sf x'=-1 \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf x'1=\frac{-2\ -\ 0}{2} \end{array}}\boxed{\begin{array}{l}\sf x''=\frac{-2}{2} \end{array}}\boxed{\begin{array}{l}\sf x''=-1 \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf S=\{-1\} \end{array}}$

Resposta correta da alternativa b) ;

$\boxed{\begin{array}{l}\sf S=\{-1\} \end{array}}$

≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡≡

$\Box \ \ \boxed{\begin{array}{l}\sf c) \ x^2+3x-10=0 \end{array}}\\\\\\\\\Box \ \ \boxed{\begin{array}{l}\sf a=1 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf b=3\end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf c=-10 \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf \Delta=b^2-4.a.c \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \Delta=(3^2)-4\ .\ 1\ .\ -10 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \Delta=9+40 \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf \Delta=49 \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf x=\frac{-b\pm\sqrt{\Delta}}{2.a} \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf x=\frac{-3\pm7}{2.1} \end{array}}\\\Box \ \ \boxed{\begin{array}{l}\sf x=\frac{-3\pm 7}{2} \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf x'=\frac{-3+7}{2} \end{array}}\boxed{\begin{array}{l}\sf x'=\frac{4}{2} \end{array}}\boxed{\begin{array}{l}\sf x'=2 \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf x''=\frac{-3-7}{2} \end{array}}\boxed{\begin{array}{l}\sf x''=\frac{-10}{2} \end{array}}\boxed{\begin{array}{l}\sf x''=-5 \end{array}}$

$\Box \ \ \boxed{\begin{array}{l}\sf S=\{-5\ ,2\} \end{array}}$

Resposta correta da alternativa c) ;

$\boxed{\begin{array}{l}\sf S=\{-5\ ,2\} \end{array}}$

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