CPhill

\((3-\sqrt{5})(4+\sqrt{5})(1+6\sqrt{5}) \)

First two products =  7 -sqrt 5

So (7 -sqrt 5) (1 + 6sqrt 5)  =

7 -sqrt 5 + 42sqrt 5 - 6*5  =

-23  + 41sqrt 5

a + b =   18

Here's a similar prob

https://web2.0calc.com/questions/truck-problem

If you get stuck, let me  know

\(9^x - 2 \cdot 3^x + 1 \)

(3^x - 1)^2

Take the derivative and  set to 0    {close your eyes if you haven't had Calculus!!   }

2 (3^x -1) ( 3^x * log 3 ) = 0

(3^x -1) * 3^x   =  0        { 3^x  is always +]      [divide this out ]

3^x -1  = 0

3^x  =1

x = 0

The min is

9^0  - 2*3^0 + 1    =     1  - 2(1) + 1   =  0

x^2 + 3y^2   =18

Take the derivative with respect to x and  set to 0

2x = 0

Take the derivative with respect to y and set to 0

6y = 0

This implies that   2x  = 6y

y = (1/3)x

So

x^2  + 3 [(1/3) x]^2  = 18

(4/3]x^2 =18

x^2  = 54/4 

x =  sqrt [54/4]    =  sqrt [27/2]  =  3sqrt (3)  /sqrt (2)  = (3/2)sqrt 6

y = (1/2)sqrt 6

Max is

 x + y =    (3/2)sqrt (6)  + (1/2)sqrt (6)  =    2sqrt 6

Area of  square  (without proof)   =   [ product of leg lengths / sum of leg lengths ] ^2  =

( [ 1 * 3 ] / [ 1 + 3 ] )^2  =  ( 3 / 4)^2  =  9 /16

I can easily prove this if you want me to 

BAC  = 60

So angle BAD  =30    (since AD is a  bisector)

Angle ABC  =45

Therefore

DB / sin BAD  = AD /sinBAD  

DB / sin 30  = 24 / sin 45

DB  / (1/2)  = 24 /sin45

DB = (1/2) * 24 / [sqrt (2) / 2)   = 12sqrt 2

Angle ADB  =  180  - 45 - 30  =  105

So

AB / sin ADB  = AD /sin ABC

AB / sin 105  = 24 / sqrt (2) /2

AB  = 24sin105 /[sqrt (2) /2 ]   = 12 + 12 sqrt (3)

AC / sin ABC  = AB /sin ACB

AC / (sqrt (2)  / 2) = (12 + 12 sqrt (3)) / sin 75

AC = (sqrt (2)  / 2) (12 + 12sqrt (3)) / sin 75   = 24

[ABC ] = (1/2) AB * AC * sin 60

[ABC ] = (1/2) (12 + 12 sqrt (3)) *24 * sqrt (3) /2 =  72 (3 + sqrt 3)  ≈ 340.7

Law of Cosines

BC  =sqrt [BD^2 + DC^2 -2 (BD * dC) cos 60]

BC =sqrt [ 4^2 + 5^2 - 2(4*5) * (1/2) ]  = sqrt 21

Law of Sines

sin BCD / BD = sin BDC / BC

sin BCD =  BD * sin BDC  /BC

xin BCD = 4 sqrt (3)  /2 / sqrt 21

sin BCD  = 2sqrt (3) / sqrt (21)  =  2/sqrt 7

Since BCD and ACE are complementary, then sin BCD = cos ACE =  2/sqrt 7

sin ACE =  sqrt [ 7 - 4] / sqrt 7  = sqrt (3)/sqrt (7)

sin AEC / sin ACE  = AC / AE

sqrt (3) / 2  / [ sqrt (3)/sqrt(7)  = sqrt (7) / 2  = AC /AE

AC = (sqrt (7) / 2) AE

Law of Cosines

AE^2   = AC^2 + CE^2   -2 ( AC * CE) cos ACE

AE^2  = (7/4)AE^2 + 4^2  - 2 ( sqrt (7) / 2 * AE * 4) (sqrt (3)  /sqrt (7)

Let AE = x

x^2  = (7/4)x^2 + 16  - 4sqrt (3) x

(3/4)x^2 - 4sqrt(3)x + 16  =  0

3x^2 - 16sqrt (3)x + 64  = 0 

Solving for x

x = 8 / sqrt 3  =   AE 

Because PX is a bisector....then PR / PQ  = RX / QX

10 / 9  = RX / QX

So RX = (10/19) ( 17)  = 170 / 19

By the Law of Cosines

PQ^2 = QR^2  + PR^2 - 2 (QR * PR) cos (YRX)

9^2 = 17^2 + 10^2 - 2(17 * 10) cos (YRX)

[ 9^2 -17^2 -10^2 ] / [ -2 * 17*10 ] = cos (YRX) =77/85

sin YRX =  sqrt [ 85^2 - 77^2 ] / 85  =  36/85

Law of Sines

XY / sin (YRX)  = RX / sin 90

XY / (36/85) = 170/19

XY = (36/85)(170/19) = 72/19  ≈  3.79

The vertex is (0,0) = (h,k)

Using the vertex form  y =a(x -h)^2 + k  and (2,8)  we can  find "a"

8 = a (2 -0)^2  + 0

8 = 4a

a = 8/4 =  2

So

y = 2 (x)^2 + 0

y = 2x^2  

a = 2   b = 0    c = 0

a + b + c  =  2

Let P  = (x,y)

We have

(x-7)^2  + (y + 11)^2  + (x -10)^2 + (y -13)^2  + (x -18)^2 + ( y + 22)^2 = 3PQ^2 + k

Simplifying the left side we have

3 x^2 - 70 x + 3 y^2 + 40 y + 1247  =   3PO^2 + k

Complete the square on x, y

3 (x^2 - 70/3x + 4900/36 x) + 3 (y^2 + 40/3 x + 1600/36)  +1247 - 4900/12 - 1600/12 =  3PQ^2 + k

3 [(x - 70/6)^2  + (y + 40/3)^2]  + 2116/3 = 3PQ^2 + k 

Q =  (70/6, -40/3) 

k = 2116/3