Question

(x + 2).(x - 8) + (x + 4)^2 = 0

Answer 1

Oie, Td Bom?!

$(x + 2) \: . \: (x - 8) + (x + 4) {}^{2} = 0 \\ x {}^{2} - 8x + 2x - 16 + x {}^{2} + 8x + 16 = 0 \\ 2x {}^{2} + 2x = 0$

• $ax {}^{2} + bx + c = 0⟶x = \frac{ - b± \sqrt{b {}^{2} - 4ac} }{2a}$

$x = \frac{ - 2± \sqrt{2 {}^{2} - 4 \: . \: 2 \: . \: 0} }{2 \: . \: 2} \\ x = \frac{ - 2± \sqrt{2 {}^{2} - 0} }{4} \\ x = \frac{ - 2± \sqrt{2 {}^{2} } }{4} \\ x = \frac{ - 2±2}{4}$

$x = \frac{ - 2 + 2}{4} = \frac{0}{4} = 0 \\ x = \frac{ - 2 - 2}{4} = \frac{ - 4}{4} = - 1$

S = $\left \{ x_{1} = - 1 \: , \: x_{2} = 0 \right \}$

Att. Makaveli1996