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.