Question

Resolver as equações de segundo grau A) X2 + x – 20 = 0 B) X2 – 81 =0 C) 5x2 + x = 0 D) 4X2 – 64 =0 E) 3x2 + 6x=0

Answer 1

$a) \: x2 + x - 20 = 0 \\ 2x = 20 \\ x = 20 \div 2 \\ x = 10$

$b) \: x2 – 81 =0 \\ x2 = 81 \\ x = 81 \div 2 \\ x = 40,5$

$c)5 \times 2 + x = 0 \\ 5 \times 2 = - x \\ - x = 10 \times ( - 1) \\ x = - 10$

$d)4x2 – 64 =0 \\ 4x = 2 + 64 \\ x = 66 \div 4 \\ x = 16$

$E) 3 \times 2 + 6x=0 \\ - 6x = 3 \times 2 \\ - 6x = 6 \\ - x = 6 \div 6 \\ - x = 1 \times ( - 1) \\ x = - 1$

Answer 2

Oie, Td Bom?!

A)

$x {}^{2} + x - 20 = 0$

$x {}^{2} + 5x - 4x - 20 = 0$

$x \: . \: (x + 5) - 4(x + 5) = 0$

$(x + 5) \: . \: (x - 4) = 0$

• $x + 5 = 0⇒x = - 5$

• $x -4 = 0⇒x = 4$

$S = \left \{ - 5 \: ,\: 4 \right \}$

B)

$x {}^{2} - 81 = 0$

$x {}^{2} = 81$

$x = ± \sqrt{81}$

$x = ±9$

$S = \left \{ - 9 \:, \: 9 \right \}$

C)

$5x {}^{2} + x = 0$

$x \: . \: (5x + 1) = 0$

• $x = 0$

• $5x + 1 = 0⇒x = - \frac{1}{5}$

$S = \left \{ - \frac{1}{5} \: ,\: 0 \right \}$

D)

$4x {}^{2} - 64 = 0$

$x {}^{2} - 16 = 0$

$x {}^{2} = 16$

$x = ± \sqrt{16}$

$x = ±4$

$S = \left \{ - 4 \: ,\: 4 \right \}$

E)

$3x {}^{2} + 6x = 0$

$3x \: . \: (x + 2) = 0$

$x \: . \: (x + 2) = 0$

• $x = 0$

• $x + 2 = 0⇒x = - 2$

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

Att. Makaveli1996