Question

equação de segundo grau
b)x(x-4)=2
c)(x+1)²=7+×​

Answer 1

Resposta:

b)

$S=\{2\pm\sqrt{6}$

c)

$S=\{-3; 2\}$

Explicação passo-a-passo:

b)

$x(x-4)=2\\x^2-4x=2\\x^2-4x-2=0$

Resolvendo por bhaskara...

$\Delta = b^2-4ac\\\Delta = (-4)^2-4\cdot 1\cdot (-2)\\\Delta = 16+8\\\\x=\frac{-b\pm\sqrt{\Delta}}{2a}\\x=\frac{-(-4)\pm\sqrt{24}}{2\cdot 1}\\x=\frac{4\pm 2\sqrt{6}}{2}\\x=2\pm\sqrt{6}$

c)

$(x+1)^2=7+x\\x^2+2x+1=7+x\\x^2+x-6=0$

Resolvendo, mais uma vez, por bhaskara...

$\Delta=b^2-4ac\\\Delta=1^2-4\cdot1\cdot(-6)\\\Delta=25\\\\x=\frac{-b\pm\sqrt{\Delta}}{2a}\\x=\frac{-1\pm\sqrt{25}}{2\cdot1}\\\\x=\frac{-1\pm5}{2}\\x_1=\frac{-1-5}{2}\rightarrow x_1=-\frac{6}{2}\rightarrow x_1=-3\\x_2=\frac{-1+5}{2}\rightarrow x_2=\frac{4}{2}\rightarrow x_2=2$

Espero ter ajudado :)