For A, is it correct that the roots/x-intercepts are the positive and negative square root of 7?For B, how could I prove that the tangents are parallel? The wording reminds me of the Mean Value Theorem, but we didn't spend a lot of time on that in class and I'm not terribly familiar with it.

A. The x-intercept is where y=0.
x^2 + 0 + 0 = 7
So yes, x = +/-sqrt(7)
B. The equation is an eclipse.
+/- sqrt(7) = +/- 2.64....
mean value theorem will tell you its continuous and differentiable, Use points on either side of +/- 2.64
(2,1) (sqrt7, 0) (3,-1)
slope: (-1-1)/(3-2) = -2
(-2,-1) (-sqrt7,0) (-3,1)
slope: (1+1)/(-3+2) = -2
If this is for calculus, take derivative at (0,sqrt7)
2x + y + x(dy/dx) + 2y(dy/dx) = 0
2(sqrt7) + 0 + (sqrt7)(dy/dx) + 0 = 0
-2 = dy/dx
likewise for x = -sqrt7
dy/dx = -2

or use the limit definition of derivative with h, as h --> zero. give tangent at specific point.