Question

as soluções da equação (×+3)(2×-4) questão (a) 2 e 3 (b) - 3 e 4 (c) 3 e 4 (d) -3 e 2​

Answer 1

Resposta:

Alternativa d)

Explicação passo-a-passo:

(x+3)(2x-4)=0

2x²-4x+6x-12=0

2x²+2x-12=0 (÷2)

x²+x-6=0

$Aplicando~a~f\'{o}rmula~de~Bhaskara~para~x^{2}+1x-6=0~~\\e~comparando~com~(a)x^{2}+(b)x+(c)=0,~temos~a=1{;}~b=1~e~c=-6\\\\\Delta=(b)^{2}-4(a)(c)=(1)^{2}-4(1)(-6)=1-(-24)=25\\\\x^{'}=\frac{-(b)-\sqrt{\Delta}}{2(a)}=\frac{-(1)-\sqrt{25}}{2(1)}=\frac{-1-5}{2}=\frac{-6}{2}=-3\\\\x^{''}=\frac{-(b)+\sqrt{\Delta}}{2(a)}=\frac{-(1)+\sqrt{25}}{2(1)}=\frac{-1+5}{2}=\frac{4}{2}=2\\\\S=\{-3,~2\}$