for which numbers $ \alpha $ is it true that $ x^2 + \alpha xy + y^2 0 $ whenever $ x $ and $ y $ are not both 0?

Question

for which numbers $ \alpha $ is it true that $ x^2 + \alpha xy + y^2 > 0 $ whenever $ x $ and $ y $ are not both 0?

jamesj · Answer

Seriously, you should try and figure it out.  Or are you truly completely stuck?

anonymous · Answer

I think we need to use the $ ax^2 + bx + c $ equation

jamesj · Answer

you can.  or you can use all of the practice you just gained on completing the square.

jamesj · Answer

what happens for instance when alpha = 2?

anonymous · Answer

when alpha = 2, $$(x + y)^2 is >= 0$$

jamesj · Answer

try some experiments.  Whan happens when
alpha = 2, 1 or 0?

jamesj · Answer

so obviously alpha = 2 doesn't work.

anonymous · Answer

if alpha is 0 the expression is also > 0

anonymous · Answer

if alpha is -2 then the expression is > 0

jamesj · Answer

is it?  I don't think so.

anonymous · Answer

when alpha is -2 then it becomes like $ x^2 - 2xy + y^2 $ which becomes $ (x-y)^2 $which is positive (when not both x and y are 0)

jamesj · Answer

but is zero whenever x = y, for example x = y = 1.

anonymous · Answer

right.

anonymous · Answer

so it's not true at alpha = -2

jamesj · Answer

or alpha = 2.

anonymous · Answer

but it's true at 0

jamesj · Answer

try completing the square and see what happens.  You know what you want: the sum of positive multiples of perfect squares

anonymous · Answer

$ x^2 + xy + y^2 + xy - xy = (x + y)^2 - xy$

anonymous · Answer

this is when alpha is 1 

if y = -x, then the thing in parentheses is 0 while the remaining term is positive since -x * -x = x^2

anonymous · Answer

I think same thing with alpha = -1

jamesj · Answer

When alpha = 1, write a for alpha
x^2 + axy + y^2 = (x + y/2)^2 + (3/4).y^2

anonymous · Answer

alpha = 0?

jamesj · Answer

agd: try completing the square for the general case

anonymous · Answer

when alpha = b ?

jamesj · Answer

No.  If I said to you: complete the square of this expression, x^2 + axy
what would you do?

anonymous · Answer

Add y^2/4 and subtract y^2/4

anonymous · Answer

yeah I will do what GT said

jamesj · Answer

a^2y^2/4 yes, to obtain what expression?

anonymous · Answer

Actually a^2 to it.

anonymous · Answer

a positive expression :-D

anonymous · Answer

oh! the discriminant!

anonymous · Answer

(x+ay/2)^2 - a^2y^2/4

jamesj · Answer

wrong.  It's not positive in general.

x^2 + axy = (x+ay/2)^2 - a^2y^2/4

jamesj · Answer

right GT.

Now, complete the square of the expression
x^2 + axy + y^2
for general a.

anonymous · Answer

$$(x^2+a^2y^2/2)^2 + y^2 - a^2y^2/2  = y^2(1 - a^2/2)    $$

anonymous · Answer

now it's clearer

jamesj · Answer

what you just wrote down isn't right, you know that, yes?

anonymous · Answer

yeah I wrote something wrong

anonymous · Answer

(x+y)^2+xy(a-2)

jamesj · Answer

$$ x^2 + axy + y^2 = (x+ay/2)^2 + (1-a^2/4)y^2 $$
Now when is that positive?

anonymous · Answer

that's positive if and only if 1- a^2/4 is positive

anonymous · Answer

and y is not 0

jamesj · Answer

yes.  We adopt everywhere to hypothesis that x and y are not both zero.

And for what values of a is that true?

anonymous · Answer

-2 < a < 2

anonymous · Answer

thanks!

jamesj · Answer

yes

anonymous · Answer

It takes me half an hour to do these questions :(

jamesj · Answer

that's why it's called learning.  This isn't 8th grade any more.

anonymous · Answer

right :-D

anonymous · Answer

I'll try the next one on my own and just post my work so folks can check if I messed something up.

jamesj · Answer

Not studying for the AI final?

anonymous · Answer

I still didn't watch the Natural Language Processing videos yet :(