Question

calcule as raízes da equação

Answer 1

11. 
a)
5x² - 20 = 0
Usando BHASKARA 

$x= \frac{-b+- \sqrt{b^2-4ac} }{2a}$

onde 
a=5
b=0
c=-20 

x= \frac{-0+- \sqrt{0^2-4.5.-20} }{2.5} 

$x= \frac{+- \sqrt{400} }{10}$

$x= \frac{+- 20 }{10}$

x= -20/10 = -2
x=20/10 = 2

b)$\frac{ \sqrt{2} }{3} x^2=0$

$x= \frac{-b+- \sqrt{b^2-4ac} }{2a}$$x= \frac{-0+- \sqrt{0^2-4 \frac{ \sqrt{2} }{3} 0} }{2 \frac{ \sqrt{2} }{3} }$

S= {0}

c) x²+x/5=0
$x^2 + \frac{1}{5}x = 0$
Usando BHASKARA 

$x= \frac{-b+- \sqrt{b^2-4ac} }{2a}$

a=1
b=1/5
c=0

$x= \frac{-1/5+- \sqrt{(1/5)^2-4.1.0} }{2.1}$

$x= \frac{-1/5+- \sqrt{(1/25)} }{2}$

$x= \frac{-1/5+- {(1/5)} }{2}$

$x'= \frac{-1/5+ {(1/5)} }{2}$

$[tex]x'= \frac{0}{2} = 0$


$x''= \frac{-1/5- {(1/5)} }{2}$

$[tex]x''= \frac{-2/5}{2}$

$[tex]x''= \frac{-2}{5} \frac{1}{2} = \frac{-2}{10} = \frac{-1}{5}$