Question

resolva a equação
a) (x+1)²=7+x

Answer 1

$(x+1)^{2}=7+x \\ x^{2} +2*x*1+1^{2} = 7+x \\ x^{2} + 2x + 1 = 7 + x \\ x^{2} + 2x - x +1 - 7 = 0 \\x^{2} + x - 6 = 0 \\ \\ D = b^2 - 4*a*c \\D = 1^2 + 4*1*6 \\D = 1+24\\D=25 \\ \\ x^{'} = \frac{-b+ \sqrt{D}}{2*a} = \frac{-1+ \sqrt{25}}{2*1} = \frac{-1+ 5}{2} = \frac{4}{2} = 2 \\ \\ x^{''} = \frac{-b- \sqrt{D}}{2*a} = \frac{-1- \sqrt{25}}{2*1} = \frac{-1- 5}{2} = \frac{-6}{2} = -3$

Portanto, $x$ possui duas raízes reais: $(-3,2)$

Answer 2

Boa tarde 

(x + 1)² = x + 7 

x² + 2x + 1 = x + 7

x² + x - 6 = 0

delta
d²= 1 + 24 = 25
d = 5

x1 = (-1 + 5)/2 = 4/2 = 2
x2 = (-1 - 5)/2 = -6/2 = -3