• Matéria: Matemática
  • Autor: viccosta
  • Perguntado 9 anos atrás

Resolva as seguintes equações
do 2°grau, identifique os coeficientes e determine as raízes se existir.

a x²-5x+6=0         


b
x²-8x+12=0     


c
x²+2x-8=0 


d
x²-5x+8=0 


e
2x²-8x+8=0 


f
x²-4x-5=0 


g
-x²-4x-5=0 


i
-x²+x+12=0 


j
6x²+x-1=0 


k
3x²-7x+2=0

Respostas

respondido por: Fabita
607
Antes de começar: \delta=b^2-4acx= \frac{-b+- \sqrt{\delta} }{2a}

a) x²-5x+6=0          
a=1,b=-5,c=6 \\ \delta =b^2-4ac=(-5)^2-4.1.6=25-24=> \delta =1 \\ x= \frac{-b+- \sqrt{\delta} }{2a} \\ x= \frac{5+- \sqrt{1} }{2.1} = \left \{ {{x= \frac{5+1}{2} = \frac{6}{2}=3 } \atop {x= { \frac{5-1}{2} = \frac{4}{2}=2 } }} \right.  \\ x=(2,3)

b) x²-8x+12=0      
a=1,b=-8,c=12 \\ \delta =b^2-4ac=(-8)^2-4.1.12=64-48=> \delta =16 \\ x= \frac{-b+- \sqrt{\delta} }{2a} \\ x= \frac{8+- \sqrt{16} }{2.1} = \left \{ {{x= \frac{8+4}{2} = \frac{12}{2}=6 } \atop {x= { \frac{8-4}{2} = \frac{4}{2}=2 } }} \right.  \\ x=(2,6)

c) x²+2x-8=0 
a=1,b=2,c=-8 \\ \delta =b^2-4ac=2^2-4.1.(-8)=4+32=> \delta =36 =>  \sqrt{\delta} = 6 \\ x= \frac{-b+- \sqrt{\delta} }{2a} \\ x= \frac{-2+- 6 }{2.1} = \left \{ {{x= \frac{-2-6}{2} = \frac{-8}{2}=-4 } \atop {x= { \frac{-2+6}{2} = \frac{4}{2}=2 } }} \right.  \\ x=(-4,2)

d) x²-5x+8=0 
a=1,b=-5,c=8 \\ \delta =b^2-4ac=(-5)^2-4.1.8=25-32=-8 \\ \delta<0
Não existem raízes reais.

e) 2x²-8x+8=0 
a=2,b=-8,c=8 \\ \delta =b^2-4ac=(-8)^2-4.2.8=64-64=> \delta =0 =>  \sqrt{\delta}=0 \\ x= \frac{-b+- \sqrt{\delta} }{2a} \\ x= \frac{8+- \sqrt{0} }{2.2} = \left \{ {{x= \frac{8+0}{4} = \frac{8}{4}=2 } \atop {x= { \frac{8-0}{4} = \frac{8}{4}=2 } }} \right.  \\ x=(2)

f) x²-4x-5=0 
a=1,b=-4,c=-5 \\ \delta =b^2-4ac=(-4)^2-4.1.(-5)=16+20=> \delta =36=> \sqrt{\delta}=6  \\ x= \frac{-b+- \sqrt{\delta} }{2a} \\ x= \frac{4+- 6 }{2.1} = \left \{ {{x= \frac{4+6}{2} = \frac{10}{2}=5 } \atop {x= { \frac{4-6}{2} = \frac{-2}{2}=-1 } }} \right.  \\ x=(-1,2)

g) -x²-4x-5=0 
a=-1,b=-4,c=-5 \\ \delta =b^2-4ac=(-4)^2-4.(-1).(-5)=16-20=> \delta =-4 \\ \delta<0
Não existem raízes reais

i) -x²+x+12=0 
a=-1,b=1,c=12 \\ \delta =b^2-4ac=1^2-4.(-1).12=16+48=> \delta =64 => \sqrt{\delta}=8  \\ x= \frac{-b+- \sqrt{\delta} }{2a} \\ x= \frac{-1+- 8 }{2.(-1)} = \left \{ {{x= \frac{-1-8}{-2} = \frac{-9}{-2}=4,5 } \atop {x= { \frac{-1+8}{-2} = \frac{7}{-2}=-3,5 } }} \right.  \\ x=(-3,5;4,5) \\ x=( \frac{-7}{2} , \frac{9}{2} )

j) 6x²+x-1=0 
a=6,b=1,c=-1 \\ \delta =b^2-4ac=1^2-4.6.(-1)=1+25=> \delta =25=>\delta= \sqrt{25}=5  \\ x= \frac{-b+- \sqrt{\delta} }{2a} \\ x= \frac{-1+- 5 }{2.6} = \left \{ {{x= \frac{-1-5}{12} = \frac{-6}{12}= \frac{-1}{2} } \atop {x= { \frac{-1+5}{12} = \frac{4}{12}=  \frac{1}{3} } }} \right.  \\ x=( \frac{-1}{2} , \frac{1}{3} )

k) 3x²-7x+2=0
a=3,b=-7,c=2 \\ \delta =b^2-4ac=(-7)^2-4.3.2=49-24=> \delta =25 => \sqrt{\delta}=5  \\ x= \frac{-b+- \sqrt{\delta} }{2a} \\ x= \frac{7+- 5}{2.3} = \left \{ {{x= \frac{7-5}{6} = \frac{2}{6}= \frac{1}{3}  } \atop {x= { \frac{7+5}{6} = \frac{12}{6}=2 } }} \right.  \\ x=( \frac{1}{3} ,2)


korvo: Fabita, para vc criar o delta no latex faça \Delta
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