Question

como se resolve a equação |3X+8:2X-3|=4

Answer 1

$3x \ + \ 8 \ \div \ 2x \ - \ 3 \ = \ 4 \\ \\ 3x \ + \ \frac{8}{2x} \ - 3 \ = \ 4 \\ \\ 6x^2 \ + \ 8 \ - \ 6x \ =\ 8x \\ \\ 6x^2 \ + \ 8 \ - \ 6x \ - \ 8x = \ 0 \\ \\ 6x^2 \ - \ 14x \ + \ 8 \ = \ 0 \\ \\ a \ = \ 6 \ \ \ \ \ \ \ \ \ \ b \ = \ - \ 14 \ \ \ \ \ \ \ \ \ \ c \ = \ 8 \\ \\ \Delta \ = \ b^2 \ - \ 4 \ * \ a \ * \ c \\ \\ \Delta \ = \ (- \ 14)^2 \ - \ 4 \ * \ 6 \ * \ 8$

$\Delta \ = \ 196 \ - \ 192 \\ \\ \Delta \ = \ 4 \\ \\ x \ = \ \frac{- \ b \ \pm \ \sqrt\ \Delta}{2 \ * \ a} \\ \\ x \ = \ \frac{- \ (- \ 14) \ \pm \ \sqrt\ 4 \ }{2 \ * \ 6} \\ \\ x \ = \ \frac{14 \ \pm \ 2}{12} \\ \\ x' \ = \ (14 \ + \ 2) \ \div \ 12 \\ \\ x' \ = \ 16 \ \div \ 12 \\ \\ x' \ = \ \frac{16}{12} \ \ = \ \ \frac{4}{3}$

$x'' \ = \ (14 \ - \ 2) \ \div \ 12 \\ \\ x'' \ = \ 12 \ \div \ 12 \\ \\ x'' \ = \ 1 \\ \\ Resposta: \ \{\frac{4}{3}; \ 1 \}$