CPhill

Here's my answer

https://web2.0calc.com/questions/number-theory-help_6

Let  S = 4   H  =  1   and  E  =   2

And we have  that

4  1  2₅

    1  2₅

______

 4  2  4₅

The  vertex form is

y  = a(x - h) ^2  +  k           where (h, k)  is  the  vertex  and (x,y) is a point on the graph

So

4 = a ( -2-  -3)^2 + 1              subtract 1 from both sides

3 = a ( 1)^2

3  = a (1)

a  = 3

So...the equation is

y = 3 ( x + 3)^2  + 1

x^4 - x^3 - 12x^2 - 3x + 1  = 0

1.  potential solutions  =   -1  or 1

2. The rational roots theorem  won't help here

We actually  have no  rational roots   

See  the (exact value) roots here calculated by WolframAlpha :

https://www.wolframalpha.com/input/?i=x%5E4+-+x%5E3+-+12x%5E2+-+3x+%2B+1%C2%A0+%3D+0

These are extremely messy......LOL!!!

We have  that

4(9) + b =  7b + 0       simplify

36 + b  =  7b       subtract b from both sides

36  =  6b      divide both sidesby 6

6  = b

4((9)  + 6    =   7(6)  +  0

36 + 6  =   42

So.....n  = 42

Area  =   (1/2) (3.14) (radius)^2

You just need to specify a radius  .........

Not the same points given, but look at my proceedure here :

https://web2.0calc.com/questions/i-don-t-understand_9

The hypotenuse of the right triangle  = sqrt  [ 20^2 + 15^2] = sqrt [ 400 + 225]  = sqrt [ 625]   = 25

So....the perimeter of the triangle is  15 + 20 + 25  =   60  in

Then....the side of the square =  60 / 4  =   15  in

And the area of the square  =   side^2  =  15^2  =  225 in^2

Thanks, HELPMEEEEEEEEEEEEE    !!!!

Here  is  another  method

If the discriminant  = 0   we  have a repeated root

(1)  (18)^2 - 4(-1)(81)  > 0      NO

(2)  (-3)^2  - 4(3)(-168) > 0    NO

(3)  (-4)^2 -  4(1) (-4) >  0    NO

(4)  (30)^2 -4(25)(9) =  900 - 900  =  0    [ this is the one with the repeated root ]

We have  the points (-1,-1)  and (1,0)  on this graph if we  had a solid line

The slope  of  this line  is  [ 0 - - 1 ] / [ 1  - -1] =  [1]/ [2]  =  1/2

So....using  the point (1,0)  and  the slope  we have either  that

y < (1/2) ( x - 1)     or    y  > (1/2)( x - 1)

Since  the point (0,0)  is in the  feasible region, note  that   0 > (1/2) ( 0 - 1)  =  0 > -1/2  is true

So....the inequality  is

y > (1/2) ( x - 1)

y >  (1/2)x  - 1/2

(1/2)x - y - 1/2 <  0            multiply through by 2

x - 2y - 1  <  0