CPhill

We have

a*0^2 + b*0 + c = 4       so  c   =  4

and

a*1^2  + b*1  + 4 =  8

a + b =  4

Since it is  monic  a = 1  and  b =3

So

f(x)  =  x^2  + 3x  + 4

If the same system = two  different  values, then we  have  no  solutions

So   K =  6

Think  about it

If  6x + 4y =  7   for some x, y

Then 6x + 4y CANNOT also =  4     for the  same  x, y   !!!!

Simplify this as

5x - 1  <  x^2 + 2x + 1 + 7x - 3

5x - 1  < x^2 + 9x - 2

0 <  x^2 - 4x - 1

So

x^2 - 4x - 1  >   0     

This is a  parabola that turns  upward......using the  Quadratic Formula, we can find  where the parabola intersects  the  x axis......the solution will lie on  (-inf , root 1) U (root 2, inf)

So

Root 1 =    [ 4 - sqrt ( (-4)^2 - 4 (1) (-1) } / 2  =  [  4 - sqrt (20) ] / 2 =   [4 - sqrt (4 * 5) ] / 2  =

[4 - 2sqrt (5) ] / 2 =  2 - sqrt 5

And by the conjugate property, Root 2 =  2 + sqrt 5

So.....the solution  is (  inf, 2 -sqrt 5)  U  ( 2 + sqrt 5, inf) 

x^2 + ?x + 10

Find all the  negative  pairs of  integers whose product = -10

(-10) (-1)  .....? =  -11

(-5)(-2)     .....?  =   -7

                sum =   -18

Note that x^2 - x - 2  can  be  factored as   (x - 2) ( x + 1)

Multiply the equation  through by this factorization and we  get

2x + 11  - x - 13 =  A(x + 1) + B ( x -2)

x - 2  = A(x + 1) +  B ( x -2)

1x - 2  = (A + B)x  + A - 2B             equting  coefficients  on  both sides  of the eqution  we have

1 = A + B

-2 =A - 2B

Multiply the second equation  through by -1

1 = A + B

2 = -A + 2B          add these

3 = 3B

So B = 1

And

A + B = 1

So A  =   0

Let the integers be

5m + 1 , 5n + 2 and  5p + 3

Their product  =    ( 5m + 1) ( 5n + 2) ( 5p + 3)  = 

(25mn + 5n + 10m + 2) ( 5p + 3)  =

125 pmn + 25pn + 50pm + 10p  + 75mn + 15n + 30m + 6

Each of the first  terms   divided by 5 will yield and integer and these added together  will also yield an integer

Only the last term  divided by 5  will yield a remainder of 1

-4 (x + 4) > x + 7

-4x  - 16 > x + 7         add 4x to  both sides, subtract 7 from  both sides

-23 > 5x                     divide both sided by 5

-23 / 5  >  x

( -inf , -23/5)

Sum of the interior angles in  a hexagon  =  (6 - 2) * 180  = 4 * 180  = 720

Angle (alpha)  =  720 - 120 -120 -120 - 130 -110  = 120°

minor arc BD = 30° + 30° =  60°

We have this relationship

Angle P = (1/2) ( minor arc BD - minor arc AC)

2 Angle P = ( 60  - minor arc AC)

And  minor arc AC =    2 * angle Q

So

2*Angle P  = 60  - 2* Angle Q

So

2 * Angle P + 2* Angle Q =  60

2 ( Angle P + Angle Q)  = 60      divide through by 2

Angle P + Angle Q  =  30°

√2  <  pi....so use the  first  function

f ( √2)  = 4 * √ 2  ≈  5.66

So

f ( f (√2)) =  f (4√2) .....now, use the last function since 4√2 ≈ 5.66

f ( 4√2) = (4√2) / 2 =   2√2