Question

Resolva, em R, a equação: log(1+x+x²)= log(2x+3)

Answer 1

$log(1+x+ x^{2} )=log(2x+3)$

CE=$\left \{ {{1+x+ x^{2} >0} \atop {2x+3>0}} \right.$


$1+x+ x^{2} =2x+3 \\ x^{2} +x-2x+1-3=0 \\ x^{2} -x-2=0$

Δ=b²-4ac
Δ=1+8
√=9
√Δ=√9

$x= \frac{-b\pm \sqrt{delta} }{2a} = \frac{-(-1)\pm \sqrt{9} }{2} = \frac{1\pm3}{2}$

$x'= \frac{1+3}{2} = \frac{4}{2} =2$

$x"= \frac{1-3}{2}=- \frac{2}{2} =-1$

Verificação

para x=2

$1+x+ x^{2} >0 \\ 1+2+2^2>0 \\ 3+4>0 \\ 7>0~~(V)$

$2x+3>0 \\ 2(2)+3>0 \\ 4+3>0 \\ 7>0~~~(V)$

para x=-1

$1+x+ x^{2} >0 \\ 1+(-1)+(-1)^2>0 \\ 1-1+1>0 \\ 1>0~~~(V)$

$2x+3>0 \\ 2(-1)+3>0 \\ -2+3>0 \\ 1>0~~~~(V)$

Logo
S={-1,2}