Question

DETERMINE AS RAÍZES REAIS DAS EQUAÇÕES ABAIXO USANDO O MÉTODO DE COMPLEMENTO DE QUADRADO.A) x² + 6x +8=0B) x² -10x -11=0C) 9x² +6x -48=0D) x² + 8x+15=0E) y² - 2y -3=0 F) x² - 14x+50=0

Answer 1

Olá, tudo bem? Vamos resolver as equações a seguir, utilizando, portanto, o MÉTODO DE COMPLEMENTO DE QUADRADO; assim:

$a)\,\,x^2 + 6x +8=0 \to x^2 + 6x +8\underbrace{+1}=0\underbrace{+1}\to \\\\x^2 + 6x+9=1\to (x+3)^2=1\to x+3=\pm\sqrt{1}\to\\\\x+3=-1\to \boxed{x=-4}\\\\ \text{ou}\\\\ x+3=1\to \boxed{x=-2}$


$b)\,\,x^2 - 10x -11=0 \to x^2 - 10x -11\underbrace{+36}=0 \underbrace{+36}\to \\\\x^2 -10x+25=36\to (x-5)^2=36\to x-5=\pm\sqrt{36}\to\\\\ x-5=-6\to \boxed{x=-1}\\\\ \text{ou}\\\\ x-5=6\to \boxed{x=11}$


$c)\,\,9x^2+6x-48=0 \to 9x^2+6x-48\underbrace{+49}=0 \underbrace{+49}\to \\\\9x^2+6x+1=49\to (3x+1)^2=49\to 3x+1=\pm\sqrt{49}\to\\\\ 3x+1=-7\to \boxed{x=-\dfrac{8}{3}}\\\\ \text{ou}\\\\ 3x+1=7\to \boxed{x=2}$


$d)\,\,x^2+8x+15=0 \to x^2+8x+15\underbrace{+1}=0 \underbrace{+1}\to \\\\x^2+8x+16=1\to (x+4)^2=1\to x+4=\pm\sqrt{1}\to\\\\ x+4=-1\to\boxed{x=-5}\\\\ \text{ou}\\\\ x+4=1\to\boxed{x=-3}$


$e) \,\,y^2-2y-3=0 \to y^2-2y-3\underbrace{+4}=0 \underbrace{+4}\to \\\\y^2-2y+1=4\to (y-1)^2=4\to y-1=\pm\sqrt{4}\to\\\\ y-1=-2\to\boxed{y=-1}\\\\ \text{ou}\\\\ y-1=2\to\boxed{y=3}$


$f) \,\,x^2-14x+50=0 \to x^2-14x+50\underbrace{-1}=0 \underbrace{-1}\to \\\\x^2-14x+49=-1\to (x-7)^2=-1\to x-7=\pm\sqrt{-1} \\\\ \texttt{Portanto, n\~ao h\'a ra\'izes reais!}$


É isso!!  :-)