Question

1- Calcule os zeros das funções dadas abaixo:
a) f(x) = x² – 5x + 6
b) f(x) = x² – 4x + 4
c) f(x) = – 2x² + 50

Answer 1

Resposta:

a) S = {3 ,2}

b) S = {2}

c) S = {5 ,-5}

Explicação passo-a-passo:

Answer 2

Oie, Td Bom?!

a)

$f(x) = x {}^{2} - 5x + 6$

$0 = x {}^{2} - 5x + 6$

$x {}^{2} - 5x + 6 = 0$

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

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

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

• $x - 2 = 0⇒x = 2$

• $x - 3 = 0⇒x = 3$

$S = \left \{ 2 \:, \: 3\right \}$

b)

$f(x) = x {}^{2} - 4x + 4$

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

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

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

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

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

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

$- (x - 2) {}^{2} = 0$

$(x - 2) {}^{2} = 0$

$x - 2 = 0$

$x = 2$

$S = \left \{ 2 \right \}$

c)

$f(x) = - 2x {}^{2} + 50$

$0 = - 2x {}^{2} + 50$

$2x {}^{2} = 50$

$x {}^{2} = \frac{50}{2}$

$x {}^{2} = 25$

$x = ± \sqrt{25}$

$x = ±5$

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

Att. Makaveli1996