Question

alguém me ajude,se ajudar bote os calculos sem fumula bhakara

$a)\frac{4}{3}x^{2}=0$
$b)x^{2}-625=0$
$c)3x^{2}+12=0$
$d)2x^{2}+108x=0$
$e)\sqrt[3]{34}x=0$
$f)4x^{2}+64x=0$
$g)2x^{2}+24x=0$
$h)4x^{2}-1600=0$

Answer 1

Resposta:

a) (4/3)*x²=0  ==>x²=0   ==> x=0

b) x²=625  ==> x=±√625  ==>x=25  ou x=-25

c)3x²=-12  ==> x²=-4     ...não existe raiz  Real

d) 2x*(x+54)=0

2x=0 ==>x=0

x+54=0  ==>x=-54

e)∛34x = 0  ==> 34x=0  ==>x=0

f) 4x*(x+16)=0

4x=0 ==>x=0

x+16=0  ==>x=-16

g)

2x*(x+12)=0

2x=0  ==>x=0

x+12=0 ==>x=-12

h)

4x*(x-400)=0

4x=0 ==>x=0

x-400=0  ==>x=400

Answer 2

a) 4x²/3 = 0

4x² = 3*0

x² = 0/4

x = √0

x = 0

===

b) x² - 625 = 0

x² = 625

x = √625

x = ±25

====

c) 3x² + 12 = 0

3x² = -12

x² = -12/3

x² = -4

x = √-4

x = ±2i

====

d)2x² + 108x = 0

x (2x + 108) = 0

x' = 0

2x = -108

x" = -54

=====

e)³√34x = 0

34x = 0³

x = 0/34

Não existe, infinitamente grande.

====

f) 4x² + 64x = 0

4x(x + 16) = 0

x' = 0

x+16 = 0

x" = -16

=====

g) 2x² + 24x = 0

2x(x + 12) = 0

x' = 0

x+12 = 0

x" =-12

=====

h) 4x² - 1600 = 0

x² = 1600/4

x² = 400

x = √400

x = ±20