Find the first three terms of the sequence given by the following.
1 answer:
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Erm
x+y=6
xy=12
x+y=6
y=6-x
sub back
xy=12
x(6-x)=12
6x-x^2=12
x^2-6x=-12
x^2-6x+12=0
quadratic formula
for
ax²+bx+c=0
for
x²-6x+12=0
a=1
b=-6
c=12
and remember that √-1=i
x=3+/-i√3
the numbers are 3+i√3 and 3-i√3
no real numbers tho
x+y=6
xy=12
x+y=6
y=6-x
sub back
xy=12
x(6-x)=12
6x-x^2=12
x^2-6x=-12
x^2-6x+12=0
quadratic formula
for
ax²+bx+c=0
for
x²-6x+12=0
a=1
b=-6
c=12
and remember that √-1=i
x=3+/-i√3
the numbers are 3+i√3 and 3-i√3
no real numbers tho