Question 477245
1)  {{{x^2 + 2xy + y^2 = 1}}}
<==> {{{(x+y)^2 - 1 = 0}}}
<==> (x+y-1)(x+y + 1) = 0.

The graph is the union of the two lines y = -x + 1 and y = -x - 1.

{{{graph( 300, 300, -5, 5, -5, 5, -x-1, -x +1)}}}

2) Solve for the system y = x and {{{y = ax^2 + 6}}} such that there is only one point of intersection.
==> Let {{{ x = ax^2 + 6}}} ==> {{{ax^2 - x + 6 = 0}}}
For there to be only one point of intersection, the discriminant has to be equal to 0.
==> {{{(-1)^2 - 4*a*6 = 0}}}, or 1- 24a = 0, or {{{a = 1/24}}}.

3)  Let {{{x = e^(i*theta)}}}.  Since {{{i = e^((1/2)*i*pi)}}}, 

we have {{{(e^(i*theta))^2 = e^(2i*theta) = e^((1/2)*i*pi)}}}
Hence we can let {{{2i*theta = (i/2)*pi}}}, or
{{{theta = pi/4}}}.
thus a complex number whose square is i is {{{e^(pi/4) = cos (pi/4) + i*sin(pi/4) = sqrt(2)/2 + i*sqrt(2)/2 = (sqrt(2)/2)(1+i)}}}