6: 1) is related to the discriminant △ of the corresponding equation. When △=0, there is an intersection point, and when △< 0, there is no intersection point. △ >; There are two intersections at 0.

2) The formula =5 is unclear, so let's take it as a square, because x12+x2 2 = (x1+x2) 2-2x1* x2 = (2m-1) 2-2 (m 2

7: 1) Bring the root of the equation into the equation to get 4+2p+q+ 1=0, so q=-2p-5, so the quadratic function is y = x 2+px-2p-4.

2) The intersection problem of quadratic function is always transformed into a discriminant problem, so △ = p 2-4q = p 2+8p+20 = (p+4) 2+4 > 0 is a constant, so the equation always has two roots, so the parabola has two intersections with the X axis.

3) q=-2p-5 from the question (1), so Y = x2+px-2p-5 = (x2+2x * p/2+p2/4)-p2/4-2p-4 = (x+p/4).

And | x1-x2 | 2 = (x1+x2) 2-4x1* x2 = p2-4 (-2p-4) = p2+8p+20 = (p+4) 2+.

If and only if p=-4, the minimum value of (p+4) 2+4 is 4, and the analytical formula of parabola is y = x 2-4x+3.

8: (1) discriminant△ = (2k+1) 2-16 (k-1/2) = 4k2+4k+1-kloc-0/6k+8.

(2) Because B and C are the two roots of the equation and the triangle is an isosceles triangle, the equation can be simplified to (x-2)(x-2k+ 1)=0. Classification discussion: when b=c, that is, the equation has a root, at this time △=0, that is, k=3/2, the unique solution of the equation is 2, because

When a=b, then c=2. At this time, b=2k- 1=4, and the perimeter of the triangle is 2+4+4 =10;

When a=c, the circumference of the same triangle is 10.