CPhill

Simplify as

s^2 - 25s + 30

This is a parabolathat turns upward

s that produces the smallest value is   (- -25) / (2 * 1)  = 25/2

Smallest value is   (25/2)^2  - 25(25/2) + 30 =  -126.25

Here :   https://web2.0calc.com/questions/help-counting_19151

Draw  a  line parallel to AB  through   R

Let PE be the altitude  of triangle APB

Extend this  altitude to intersect the line we  have constructed at F

APB is a right triangle  because  8^2 + 4^2  = 80 =  4sqrt (5)

Now the area of APB =   (1/2)(4)(8) =  16

To find PE

[APB ] =  (1/2) sqrt (80) *PE

16  =(1/2) sqrt (80) * PE

32 / sqrt (80) = PE  =  32 / [ 4sqrt5 ] = 8/sqrt5 

We can find  EB  as sqrt ( PB^2 - PE^2) =    sqrt ( 4^2  - 64/5 )  = sqrt (16/5) = 4/sqrt5

And triangle RDC is  congruent to triangle APB

Let its altitude = RG = 8/sqrt5

And, by symmetry, DG = 4/sqrt5

We can  now form  right triangle PFR  with PR the hypotenuse

PF =  PE + BC + RG  =   2(8/sqrt5)  +  4sqrt (5)  =   16/sqrt 5 + 4*5/sqrt 5  =  36/sqrt 5

And FR = AB - EB - DG =  4sqrt 5 - 2(4/sqrt5) =  4*5 /sqrt 5  - 8/sqrt 5  =   12/sqrt 5

So

PR^2  =  PF^2  + FR^2

PR^2  = 1296/5 + 144/5  = 1440 / 5

PR = sqrt (1440/5) =  sqrt (288) =   12sqrt2

Simplify as

x^2 + 20x - 13 =  0

Product of the  roots = ab = -13      (ab)^2  = 169

2ab = -26

Sum of the  roots  = a + b =  -20

(a + b)^2  = (-20)^2

a^2 + 2ab + b^2  = 400

a^2 + b^2  + 2ab = 400

a^2 + b^2  - 26  = 400

a^2 + b^2 =  426

For  the new quadratic,  the  sum of the roots =  (a^2 + b^2) + (a + b) + 2 =  426 - 20 + 2  =  408

The product of the roots = (a^2 + a + 1) (b^2 + b + 1)   =

a^2 b^2 + a^2 b + a^2 + a b^2 + a b + a + b^2 + b + 1 =

(a^2 + b^2) + (ab)^2 + ab (a + b) + ab + (a + b) + 1  =

426 + 169 + (-13)(-20) -13 - 20 + 1  = 823

The   quadratic is  Ax^2 + Bx + C  and we  can let A = 1

x^2 - 408x + 823

Thx, BlackjackEd....here's another way without having to actually calculate the roots

Simplify as

x^2 - 23x - 16  =  0

Product of the roots =  ab  = -16

2ab = -32

2(a^2b^2)  =  2(ab)^2  = 2(-16)^2  = 512

Sum of the  roots =  a + b = 23

(a + b)^2  = 529

(a^2 + 2ab + b^2) = 529

(a^2 + b^2)  + 2ab = 529

(a^2 + b^2) + (-32) = 529

(a^2 + b^2) -  32 = 529

(a^2 + b^2)  = 529  + 32   =  561

(a^2 + b^2)^2  =  a^4 + 2a^2b^2  + b^4

(a^2 + b^2)^2 - 2a^2b^2  = a^4 + b^4

(561)^2  - 512  =  a^4 + b^4

314209  = a^4 + b^4

For this to be a circle, A  must =  2

So....we have

2x^2 + Bx + 2y^2 + Cy = 50     divide through by 2

x^2 + (B/2)x + y^2 +(C/2)y  = 25     complete the square on x,y

x^2 + (B/2)X + B^2/4 + y^2 + (C/2)y + C^2/4  = 25  + B^2/4 + C^2/4

Factor

(x + B/2)^2  + ( y + C/2)^2 =  25 + B^2/4 + C^2/4

Since the center is  (-5,2)  then

B/2 = 5         C/2 =  -2

B = 10          C = -4

The radius^2  =  25 + 10^2/4  + 16/4 =   25 + 25 + 4  = 54

r = sqrt (54)

A + B + C + r =   2 + 10 - 4  + sqrt (54)  = 8 + sqrt (54)

You're Welcome  !!!

Here : https://web2.0calc.com/questions/algebra_58291

4.  Simplified we get

2x^2 -kx + 23  = 0

We have no real solutions when the discriminant is <  0

So

k^2  - 4(2)(23) < 0

k^2  -  184  < 0

k^2  <  184

k <  sqrt (184)   ≈ 13.564

So  the largest integer =  13

3.   x^2 - 3x  + 5  = 0 

Note that if we  just  divide through by 5   we get

(1/5)x^2  - (3/5)x + 1  = 0

A =1/5     B  =(-3/5)

A + B  =  1/5    -3/5 =  -2/5

{You can check to see that the roots are the same }