16a+4b+c=-6

4a-2b+c=0

The solution is b=- 1-2a c=-2-8a.

The coordinate of point C is (0, -2-8a), and the height of triangle ABC at the bottom of AB is |-2-8a|=2+8a.

There must be a point (-2,0) in A and B, which is the point that intersects the X axis under the known conditions.

According to Vieta theorem x1+x2 =-b/a = 2+1/a.

So the other point is (4+ 1/a, 0).

The AB distance is 6+1/a.

So the ABC area of the triangle =1/2 * (2+8a) * (6+1/a) =10+1/a+24a ≥10+2 * √ (/kloc

Take the equal sign if and only if x=√6/ 12.

(2)[(x^2-2x)^2-3(x^2-2x)-(k^2+3k)]/(x^2-2x-2k)=0

After sorting, (x2-2x+k) (x2-2x-k-3)/(x2-2x-2k) = 0.

If four different solutions are needed, then the discriminant of the above two equations is greater than 0. You can find the range of K.

Moreover, k, -k-3 and -2k are required to be different, that is, k is not equal to 3/2 and k is not equal to 3.

(3) Extend the intersection of BG and AB to point D, where D is the midpoint of AB. And DG= 1/2*CG= 13/2.

Applying cosine theorem in triangle ADG and triangle BDG, there are

AD^2+DG^2-AG^2=2cos∠ADG*AD*DG

BD^2+DG^2-BG^2=2cos∠BDG*BD*DG

Let AD=BD=x, substitute it into the known value, and apply cos∠ADG=-2cos∠BDG.

So we can get x= 13/2.

Then AB= 13

There is ag 2+BG 2 = ab 2.

Triangle ABG is a right triangle, and the height from G to AB is 5 *12/13 = 60/13.

△ Height on AB △ABC side =3 times the height from G to AB = 180/ 13.