Question

y=(a-ib-a)2+b2
y is supposed to equal 0. how? I get it to 2b^2
i^2=-1

Answer 1

confused
is this
\[y=(a+ib-a)^2b^2\]?

Answer 2

\[y=(a+ib-a)^2+b^2\]

Answer 3

f(x)=(x-a)^2+b^2
g(x)=-(x-a)^2+b^2

Answer 4

then you get 0 for sure because
\[a-a=0\] and you have
\[y=(bi)^2+b^2=i^2b^2+b^2=-b^2+b^2=0\]

Answer 5

yes but f(x) has zeros at a+-b

Answer 6

\[f(x)=(x-a)^2+b^2\]
\[ g(x)=-(x-a)^2+b^2\]what is the question?

Answer 7

prove that f(x) has its zeros at a±ib where i=√(-1).

Answer 8

ah ok
then write
\[(x-a)^2=-b^2\] as a first step

Answer 9

then
\[x-a=\pm\sqrt{-b^2}=\pm\sqrt{-1}\sqrt{b^2}=\pm bi\]

Answer 10

and so
\[x=a\pm bi\]

Answer 11

aha makes sense

Answer 12

but then, how can i prove that f(x)+g(x)=0?

Answer 13

hmm i don't think it is zero

Answer 14

its supposed to be :(

Answer 15

\[f(x)+g(x)=(x-a)^2+b^2+-(x-a)^2+b^2=2b^2\]

Answer 16

possibly theres a complez zero at the height 2b^2

Answer 17

\[f(x)+g(x)=2b^2\] is a constant. not zero unless b is zero

Answer 18

no, there are no complex zeros.
\[f(x)+g(x)\] is a constant, there is no x in it

Answer 19

what does the question say exactly?

Answer 20

well we have a formula for quartics and one for cubics and i need to find a general statement true for both.

Answer 21

??

Answer 22

Cubics:
y1=(x+2)(x-(3+2i))(x+(3-2i))=x3-4x2+x+26
y2=-(x+2)(x-(3+2))(x-(3-2))=-x3+4x2+7x-10
ym=4x+8

y1+y2=8x+16=2(4x+8)=2ym

Answer 23

my statement is that f(x)+g(x)=2ym

Answer 24

what does ym mean?

Answer 25

thats a straight line which has an equal vertical distance to f(x) and g(x) at all points.

Answer 26

in the quartics case it is the line y=0

Answer 27

ok so in simple language y = f(x) + g(x) ?

Answer 28

well 2y=f(x)+g(x)

Answer 29

ok

Answer 30

so now you have
\[f(x)=(x-a)^2+b^2,g(x)=-(x-a)^2+b^2\] and
\[f(x)+g(x)=2b^2\] right?

Answer 31

btw these are quadratics not "quartics" because the degree is 2, not 4

Answer 32

ah i got confused, the straight line ym isnt at the height 0, its at b^2

Answer 33

ooooooooooooh i think i see
the line that is equidistant from these two is the line y = 0

Answer 34

i dont know what equidistant means unfortunately haha

Answer 35

omg hold on this is such a stupid question

Answer 36

of course f(x)+g(x)=2b^2... b^2=ym

Answer 37

they share a vertex

Answer 38

right they sure do

Answer 39

i get it now though haha thanks for your time ^^

Answer 40

and that is (a,b^2) so they are equidistant from the line
\[y=b^2\]

Answer 41

yw