Question

resolva as seguintes equações do 2º grau , identifique os coeficientes e determine as raizes se existir.
a) x²-5+6=0
b) x²+2x-8=0
c) x²-5x+8=0
d)-x²+x+12=0
e)-x²+6x-5=0
f)6x²+x-1=0
h) 2x²-7x+2=0
i) 4x²+9=12x
j) 2x²= -12x-18

Answer 1

A)x²-5+6=0
x²+1=0
x²=-1
x=√-1
∅

B)x²+2x-8=0
a=1 b=2 c=(-8)
∆=b²-4ac
∆=2²-4.1.(-8)
∆=4+32
∆=36√36=6
X= -b±√∆/2a
X= -4±6/2.1
X'= -4+6/2
X'=2/2=1
X”= -4-6/2
X"= -10/2
X"= -5
S{-5,1}

C)x²-5x+8=0
a=1 b=(-5) c=8
∆=b²-4ac
∆=(-5)²-4.1.8
∆=25-32
∆= -7√-7
∅

D) -x²+x+12=0
a=(-1) b=1 c=12
∆=b²-4ac
∆=1²-4.(-1).12
∆=1+48
∆=49√49=7
X= -b±√∆/2a
X= -1±7/2.(-1)
X'= -1+7/2
X'=6/2
X'=3
X"= -1-7/2
X"= -8/2
X"= -4
S{-4,3}

E) -x²+6x-5=0.(-1)
x²-6x+5=0
a=1 b=(-6) c=5
∆=b²-4ac
∆=(-6)²-4.1.5
∆=36-20
∆=16√16=4
X= -b±√∆/2a
X= -(-6)±4/2.1
X'=6+4/2
X'=10/2
X'=5
X"=6-4/2
X"=2/2
X"=1
S{1,5}

F)6x²+x-1=0
a=6 b=1 c=(-1)
∆=b²-4ac
∆=1²-4.6.(-1)
∆=1+24
∆=25√25=5
X= -b±√∆/2a
X= -1±5/2.6
X= -1±5/12
X'= -1+5/12
X'=4/12=4:4=1 12:4=3
X'=1/3
X"= -1-5/12
X"= -6/12= -6:3= -2 12:3=4
X"=-2/4
S{'2/4,1/3}

H)2x²-7x+2=0
a=2 b=(-7) c=2
∆=b²-4ac
∆=(-7)-4.2.2
∆=49-16
∆=33√33
∅

I)4x²+9=12x
4x²-12x+9=0
a=4 b=(-12) c=9
∆=b²-4ac
∆=(-12)-4.4.9
∆=144-144
∆=0√0=0
X= -b±√∆/2a
X= -(-12)±0/2.2
X'=X"=12/4=3
S{3}

J)2x²=-12x-18
2x²+12x+18=0
a=2 b=12 c=18
∆=b²-4ac
∆=12²-4.2.18
∆=144-144
∆=0√0=0
X= -b±√∆/2a
X= -12±0/2.2
X'=X"= -12/4=±3
S{-3,3}