Question

calcule os monomios
(a-1).(a+1)
(x-2).(x+3)

Answer 1

a²+a-a-1
a²-1=0
0±√0-4.1.(-1) /2.1
0±2 /2
x' = 2/2 = 1
x" = (-2)/2 = -1

x²+3x-2x-6
x²+x-6=0
-1±√1²-4.1.(-6) /2.1
-1±√1+24 /2
-1±√25 /2
-1±5/2
x' = -1+5/2 = 4/2 = 2
x" = -1-5/2 = (-6)/2 = -3

Answer 2

(a-1).(a+1)=
a2+1a-1a-1=
a2-1=0
a= \| 1
a1= +1
a2= -1

(x-2). (x+3)=
x2+3x-2x-6=0
x2+x-6=0

-b+- \| b2-4ac /2a

-1 +- \| 1+24 /2
-1 +- \| 25 /2
X'= -1-5 /2
X'= -6/2
X'= -3

X"= -1+5 /2
X"= 4/2
X"= 2