Matemática, perguntado por evelyns1, 1 ano atrás

As raízes da equação (2x + 7) . (3x – 4)

Com foto ou cálculo explicado

Soluções para a tarefa

Respondido por caio0202
1
\mathtt{(2x + 7)~.~(3x-4) = 0} \\ \mathtt{Distributiva} \\ \\ \mathtt{6x^2-8x+21x- 28 = 0} \\ \mathtt{6x^2 + 13x -28 = 0} \\ \\ \mathtt{a = 6} \\ \mathtt{b = 13} \\ \mathtt{c = -28} \\ \\ \mathtt{\Delta = b^2-4~.~a~.~c} \\ \mathtt{\Delta = (13)^2-4~.~6~.~(-28)} \\ \mathtt{\Delta = 169 -24~.~(-28)}\\ \mathtt{\Delta =  169 + 672} \\ \mathtt{\Delta = 841}}} \\ \\ \mathtt{\dfrac{-b+-~\sqrt\Delta}{2~.~a}} ~~=~~ \mathtt{\dfrac{-13+-~\sqrt{841}}{2~.~6}}~~=~~ \mathtt{\dfrac{-13 +-~29}{12}}


\mathtt{x' = \dfrac{-13 +29}{12}~~=~~ \mathtt{\dfrac{16}{12} ~(\div4) = \dfrac{4}{3}}} \\ \\ \\ \mathtt{x'' = \dfrac{-13 -29}{12}}~~=~~ \mathtt{-\dfrac{42}{12}~~ (\div6)~=-\dfrac{7}{2}}  \\ \\ \\ \boxed{\boxed{\mathtt{Resposta : S\{\dfrac{4}{3}~~~-\dfrac{7}{2}} }}
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