Respostas

2014-05-14T05:01:45-03:00
a) 2x^2-18=0\\\\2x^2=0+18\\\\x^2= \frac{18}{2} \\\\x^2=9\\\\x=\pm \sqrt{9}
**************************************************************************************************
-x^2+4=0\\\\4=0+x^2\\\\\pm \sqrt{4} =z\\\\\pm2=x
***************************************************************************************
3x^2+21=0\\\\3x^2=-21\\\\x^2= \frac{-21}{3} \\\\x^2=-7
não existe raíz quadrada de numero negativo..
portanto o Δ dessa equação é negativo e a equção não possui raízes reais (ela nao corta o eixo x)
*********************************************************************************
2x^2+4=0\\\\2x^2=-4\\\\x^2= \frac{-4}{2}
outra vez raíz de numero negativo rs ;)
********************************************************************************
x^2+x=0
o unico jeito de isso dar 0 é quando:
x= 0
********************************************************************************
x^2+x=0
vc pode resolver por bhaskara tbm ;)
a = 1 
b = 1
c = 0
 \frac{-b\pm \sqrt{b^2-4*a*c} }{2*a} = \frac{-1\pm \sqrt{1^2-4*1*0} }{2*1} = \frac{-1\pm \sqrt{1} }{2} \\\\x'= \frac{1+1}{2} =1\\\\x''= \frac{1-1}{2} =0
*****************************************************************************************
5x^2+25x=0

simplificando dividindo por 5
x^2+5x=0
a=1 
b= 5
c= 0
\frac{-b\pm \sqrt{b^2-4*a*c} }{2*a} = \frac{-5\pm \sqrt{5^2-4*1*0} }{2*1} = \frac{-5\pm5}{2} \\\\x'= \frac{-5+5}{2} =0\\\\x''= \frac{-5-5}{2} =-5


2 5 2
2014-05-14T07:38:12-03:00
A) 2x² - 18 = 0
 2x²=18
x²=18/2
x²=9
x=√9
x=3 ou x= -3
         
b) -x²+4=0
-x²= -4
x²=4
x=√4
x=2 ou x= -2

c)3x²+21=0
3x²= -21
x²= -21/3
x² = -7
x=√-7

d) 2x²+4=0 
2x²= -4
x²= -4/2
x²= -2
x=√-2

e) X² +x =0
x(x+1)=0
x=0 ou(x+1=0) x= -1

f) 5x² +25x=0 
x(5x+25)+0
x=0 ou (5x+25=0)  5x= -25 x= -25/5  x=-5