CPhill

-1 < 2x < 12    divide through by 2

-1/2 < x < 6

And

-1 < 3x  -7 < 12       add 7

6 < 3x < 19      divide through by 3

2 < x  < 19/2 

2 < x < 6   

We have a lower degree polynomial over a higer degree polynomial....in this situation,the limit  always =  0 

x^2 + mx + 4 =  2x^2 + 17x + 8        rearrange as

x^2 + (17 - m)x  + 4  =  0

If we have two distinct roots, the discriminant > 0  ...... so....

(17 - m)^2  - 4(1)(4)  >  0

(17 - m)^2  >  16          take the positive root

17 - m   >    4

13 > m

m <  13

ab - 3a + 4b  =  131

b ( a + 4)  =  131  + 3a

b =  (131  + 3a)  / ( a + 4)

a         b

3       20

13     10

min  l a - b l  = l 13  -10 l   =  3

        A

        17

   x       x

B           D            C

17

                    E

"Line segment AD is extended to E, such that angle DBE = angle BAD = 17 degrees."  This has no bearing on the problem.  

If AD  = AB  then angle  ABD  =  angle ADB

So  angle   ABD  = (180 - 17)  / 2   =  81.5°  

A

    2

       H

           x

C     1      B

Let HB = x

AC^2 = AB^2 - BC^2  =   (2 +x)^2 - 1^2  =  x^2 + 4x + 3

CH^2  = BC^2 - BH^2 =   1 - x^2      (1)

CH^2 = AC^2 - AH^2  =  (x^2 + 4x + 3) - 2^2  = x^2 + 4x -1    (2)

Set (1)   =  (2)

1-x^2  = x^2 + 4x  -1

2x^2  + 4x -2  =  0

x^2 + 2x - 1  =  0

x^2 + 2x  = 1      complete the square on x

x^2 + 2x + 1  = 1 + 1

(x + 1)^2   = 2          take the positive root

x + 1  =sqrt (2)

x = sqrt (2)  - 1

CH^2   =  BC^2 - BH^2 =   1^2 - (sqrt 2 -1)^2   =  1 - (2 - 2sqrt 2 + 1)  =  2sqrt (2) - 2  

CH =   sqrt  (2sqrt (2)  - 2)  =  sqrt (sqrt (8) - 2)  ≈  .91

Another way  using similar triangles

Draw altitudes AM   and FN to the base DE

Triangles DAM  and DFN are similar

AD = (3/5) AF

angle DFN = 30°....DN = 2.5

angle FDN  = 60°....FN  = 2.5 sqrt (3)  = (5/2)sqrt (3)    

So   AM = (3/5)FN =   (3/5)(5/2)sqrt 3  =  (3/2)sqrt 3 =  sqrt (27/4)

DN = 5/2  

So

DM = (3/5)(5/2)  =  3/2

So  ME = DE - DM =    5 - 3/2 =  7/2

So

AE  = sqrt [ ME^2 + AM^2 ]  = sqrt [ (7/2)^2 + 27/4 ]  =  sqrt [ 49 + 27 ] / 2  = sqrt [ 76] /2  = 

sqrt [ 19 * 4 ]  / 2   =  2 sqrt [ 19]   / 2  =   sqrt (19)

We can  solve this with the Law of Cosines

AF =2   FE  =5    angle DFE = 60

cos (DFE) =cos (60)  =1/2

So

AE^2  = AF^2 + FE^2  - 2 (AF *FE) cos (60)

AE^2  = 2^2 + 5^2 - 2(2 * 5) (1/2)

AE^2 = 29 - 10

AE^2  = 19

AE = sqrt (19) ≈ 4.36

    Q

                    5     

3           X      

    P                 R

               4

Y = P

Angle QPR  = 90°

sin angle (QRP) = (3/5)

Angle RPX  = 45°

sin angle RPX = 1/sqrt 2

Because PX is an angle bisector

QP / QX  = PR / RX

Let RX = x     QX =  5 - x

So

3/ (5-x)  =  4 / x

3x = 4 (5-x)

3x = 20 - 4x

7x = 20

x = 20/7

Law of Sines

(YX)  /sin (QRP)  = RX / sin (RPX)

YX / (3/5)  = (20/7) / ( 1/sqrt 2)

YX  =  (3/5)(20/7) * sqrt 2  =   (60/35)sqrt 2  =  (12/7)sqrt (2) ≈ 2.42

(2^1 - 3) + (2^2 -3) +(2^3 - 3) + (2^4 -3) + (2^5 - 3)  =

-1 + 1   + 5  + 13 + 29  = 

47