Question

Me ajudem sem usar baskara por favor preciso de entregar esse trabalho daqui a pouco √x²+3x+2=2√3

Answer 1

$\\ ( \sqrt{x^2+3x+2})^2= (2\sqrt{3})^2\\ x^2+3x+2= 4\cdot3\\ x^2+3x+2=12\\ x^2+3x+2-12=0\\ x^2+3x-10=0$

Fazendo a prova real temos:
Se x = -5
$\sqrt{x^2+3x+2}= 2\sqrt{3} \\ \\\sqrt{(-5)^2+3\cdot(-5)+2}= 2\sqrt{3} \\ \\\sqrt{25-15+2}= 2\sqrt{3} \\ \\\sqrt{12}= 2\sqrt{3} \\ \\\sqrt{4\cdot3}= 2\sqrt{3} \\ \\\sqrt{4}\cdot\sqrt{3}= 2\sqrt{3} \\ \\2\sqrt{3} = 2\sqrt{3}$

Se x = -5
$\sqrt{x^2+3x+2}= 2\sqrt{3} \\ \\\sqrt{2^2+3\cdot2+2}= 2\sqrt{3} \\ \\\sqrt{4+6+2}= 2\sqrt{3} \\ \\\sqrt{12}= 2\sqrt{3} \\ \\\sqrt{4\cdot3}= 2\sqrt{3} \\ \\\sqrt{4}\cdot\sqrt{3}= 2\sqrt{3} \\ \\2\sqrt{3} = 2\sqrt{3}$
logo, ambas raízes são solução.
S = {-5, 2}