CPhill

x^2 - 6x + c < 0

Set this  = 0

If this has real solutions, then the discriminant  is  ≥ 0

So

6^2 - 4c ≥ 0

36 - 4c ≥ 0

36 ≥ 4c

c ≤ 9

When c  = 9...the parabola will have its vertex at (3,0)......so....since it turns upward, this is the low point on the graph....so...it is not less than 0  at any point

So.....this implies  that  the inequality will be < 0 [ i.e., have real solutions for x ]   when c  = (0, 9)

Here's a graph to show this : https://www.desmos.com/calculator/frqkypfzn8

If the number doubles every two years, then we have five doublings  from 1990 to 2000....so we have

1,000,000 (2)^5  =  

32,000,000  in year 2000

OOPS!!!....thx, guest!!!....correction made

LOL!!!!!

The x coordinate of the vertex  can be found as   -b / (2a)  =  2   ⇒  -b = 4a  ⇒  b  = -4a

So.....we have the following quadratic

y = ax^2 - 4ax + c

And since  (2,1)   and (-4, -3)  are on the graph, we have that

a(2)^2 - 4a(2) + c  =  1  ⇒   4a - 8a + c  = 1  ⇒    -4a + c  =  1   ⇒   -32a + 8c  = 8    (1) 

a(-4)^2  - 4a(-4) + c  = -3  ⇒   16a + 16a + c  = -3  ⇒  32a + c  = -3      (2)

Add (1)  and (2)   and we have that

9c = 5

c = 5/9

And

-4a + 5/9  =  1

-4a =  4/9

a  = -1//9

And b  = -4(-1/9)  = 4/9

So....the quadratic  is

y  = (-1/9)x^2 + (4/9)x + 5/9

Setting this to 0  and we have that

(-1/9)x^2  + (4/9)x + (5/9)  =  0         multiply through by  -9

x^2 - 4x - 5  = 0       factor

(x - 5) ( x + 1)  = 0      set each factor to 0 and solve for x and we have that

x - 5  = 0                x + 1  = 0

x  = 5  = m             x  = -1 = n

So....m - n  =   5 -(-1)   =   6

Let D be any point in the interior of ABC.....draw BD  and extend this to E, lying on AC

So.....

DE + EC > DC      ....  add BD to both sides

(DE + BD) + EC > DC + BD

BE + EC > DC + BD

AB + AE > BE .......add EC to both sides

AB + (AE + EC) > BE + EC

AB + AC > BE +EC           which implies that

AB + AC > DC + BD

The losing candidate could actually have received as little as 30.5%  of the vote  [ this would be rounded to 31% ]

If there were 100 votes cast, the losing candidate would have received  30.5 votes  [ not possible]

But, doubling this number of votes  = 200   and 30.5%  of 200  - 61 votes

So....the least number of votes that could have been cast  = 200

There are 4 equal parts to AC   and BC is  3 of them

The coordinates of C are given by

[ (2 + 3(3- 2) , -6 + 3(-1 - (-6) ) ]  =   [ 2 + 3(1) , -6 + 3(5) ]  = [ 2 + 3, -6 + 15 ]  = (5, 9)

If (0,0)  is on the graph, then c  = 0

So.....we have these two equations

a(-3)^2 + b(-3)  =  3    ⇒ 9a - 3b  = 3      (1)

a(1)^2 + b(1)  =    3   ⇒ a  + b  = 3  ⇒  3a + 3b =  9     (2)

Add (1)  and (2)   and we have that  

12a  =  12

a  = 1

And using (1)

9)1) - 3(b)  = 3

9 - 3b  = 3

-3b  = -6

b  = 2

So   100a  + 10b + c   =    100(1) + 10(2) + 0   =   100 + 20 =  120

Recursive rule

a_n+1  =  a_n * (-4)    where  a₁  =  3