Question

1- Encontre as Raízes das equaçoes abaixo, se Existirem a) 3x2-7x=0 b) 2x2+26=0 c)-2x2+26=0 d)-5x2-35x=0

Answer 1

Explicação passo-a-passo:

a)

$\sf 3x^2-7x=0$

$\sf x\cdot(3x-7)=0$

• $\sf \red{x'=0}$

• $\sf 3x-7=0~\rightarrow~3x=7~\rightarrow~\red{x"=\dfrac{7}{3}}$

b)

$\sf 2x^2+26=0$

$\sf 2x^2=-26$

$\sf x^2=\dfrac{-26}{2}$

$\sf x^2=-13$

Não há raízes reais

c)

$\sf -2x^2+26=0$

$\sf 2x^2=26$

$\sf x^2=\dfrac{26}{2}$

$\sf x^2=13$

$\sf x=\pm\sqrt{13}$

• $\sf \green{x'=\sqrt{13}}$

• $\sf \green{x"=-\sqrt{13}}$

d)

$\sf -5x^2-35=0$

$\sf 5x^2=-35$

$\sf x^2=\dfrac{-35}{5}$

$\sf x^2=-7$

Não há raízes reais

Answer 2

Oie, Td Bom?!

a)

$3x {}^{2} - 7x = 0$

$x \: . \: (3x - 7) = 0$

______________________________________________

$x = 0$

$3x - 7 = 0⇒x = \frac{7}{3}$

________________________________________________

$S= \left \{0 \: , \: \frac{7}{3} \right \}$

b)

$2x {}^{2} + 26 = 0$

$x {}^{2} + 13 = 0$

$x {}^{2} = - 13$

$x∉\mathbb{R}$

c)

$- 2x {}^{2} + 26 = 0$

$x {}^{2} - 13 = 0$

$x {}^{2} = 13$

$x = ± \sqrt{13}$

_________________________________________________

$S= \left \{ - \sqrt{13} \: , \: \sqrt{13} \right \}$

d)

$- 5x {}^{2} - 35x = 0$

$- 5x \: . \: (x + 7) = 0$

$x \: . \: (x + 7) = 0$

_________________________________________________

$x = 0$

$x + 7 = 0⇒x = - 7$

______________________________________________,__

$S= \left \{ - 7 \:, \: 0 \right \}$

Att. Makaveli1996