Question

(3x+1)(3x-1)= 2 + x(x-2)pessoal me desculpem a questão e para ser resolvida pela formula de Bháskara

Answer 1

$(3x+1)(3x-1)=2+x(x-2)\\9x^2-1=2+x^2-2x\\8x^2+2x-3=0\\ \\\Delta=2^2-4*8*(-3)\\\Delta=4+96\\ \\\Delta=100\\ \\ \\x= \frac{-2 \pm\sqrt{100} }{2*8} \\ \\x= \frac{-2\pm10}{16}\\ \\ \\x'= \frac{-2+10}{16}\\ \\x'= \frac{1}{2}$

$x''= \frac{-2-10}{16} \\ \\x''= \frac{-12}{16} \\ \\x''=- \frac{3}{4}$

Answer 2

(3x+1)(3x-1)= 2 + x(x-2)⇒ 9x²-3x+3x-1=2+x²-2x⇒  9x²-1=2+x²-2x⇒  9x²-x²+2x-1-2=0⇒
8x²+2x-3=0
a=8; b=2; c=-3
∆=b²-4ac→ ∆=2²-4.8.(-3)⇒ Δ=4+96⇒ Δ=100
x=-b±√∆ / 2.a→ x=-2±√100 / 2.8⇒ x=-2±10 / 16
x'=-2+10 / 16⇒ x'=8/16(simplifique por 8)⇒ x'=1/2
x"=-2-10 / 16⇒ x"=-12/16(simplifique por 4)⇒ x"=-3/4
S={1/2 , -3/4}