Matemática, perguntado por juliacorreia052003, 1 ano atrás

-3x^2+8x+17=0
Equação de 2 grau

Soluções para a tarefa

Respondido por caio0202
0

Equação do 2º:

 \mathtt{-3x^2+8x+17=0} \\ \\ \mathtt{a = -3} \\ \mathtt{b=  8} \\ \mathtt{c = 17} \\ \\ \mathtt{\Delta =b^2-4~.~a~.~c} \\ \mathtt{\Delta =8^2-4~.~(-3)~.~17} \\ \mathtt{\Delta =64+12~.~17} \\ \mathtt{\Delta =64+204} \\ \mathtt{\Delta =268}

 \mathtt{\dfrac{-b+-~\sqrt{\Delta}}{2~.~a}~~=~~\dfrac{-8+-~\sqrt{268}}{2~.~(-3)}~~=~~\dfrac{-8+-2\sqrt{67}}{-6}~~~(\div2)}   \\ \\ \\ \mathtt{x' = \dfrac{-4+~\sqrt{67}}{-3}~~=~~\dfrac{4+\sqrt{67}}{3}} \\ \\ \\ \mathtt{x'' = \dfrac{-4-\sqrt{67}}{-3}~~=~~\dfrac{4-\sqrt{67}}{3}} \\ \\ \\  \boxed{\boxed{\mathtt{Resposta: ~~ x'= \dfrac{4+\sqrt{67}}{3}~~=~~x'' =\dfrac{4-\sqrt{67}}{3}}}}

Respondido por caio0202
0

Equação do 2º:

 \mathtt{-3x^2+8x+17=0} \\ \\ \mathtt{a = -3} \\ \mathtt{b=  8} \\ \mathtt{c = 17} \\ \\ \mathtt{\Delta =b^2-4~.~a~.~c} \\ \mathtt{\Delta =8^2-4~.~(-3)~.~17} \\ \mathtt{\Delta =64+12~.~17} \\ \mathtt{\Delta =64+204} \\ \mathtt{\Delta =268}

 \mathtt{\dfrac{-b+-~\sqrt{\Delta}}{2~.~a}~~=~~\dfrac{-8+-~\sqrt{268}}{2~.~(-3)}~~=~~\dfrac{-8+-2\sqrt{67}}{-6}~~~(\div2)}   \\ \\ \\ \mathtt{x' = \dfrac{-4+~\sqrt{67}}{-3}~~=~~\dfrac{4+\sqrt{67}}{3}} \\ \\ \\ \mathtt{x'' = \dfrac{-4-\sqrt{67}}{-3}~~=~~\dfrac{4-\sqrt{67}}{3}} \\ \\ \\  \boxed{\boxed{\mathtt{Resposta: ~~ x'= \dfrac{4+\sqrt{67}}{3}~~=~~x'' =\dfrac{4-\sqrt{67}}{3}}}}

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