Solutions to Let's Get Started

Solutions to problems:
1. y = 8, -2
2. 5050
3. f'(x) = 3 * ln(2x^3) + 9

1.          y^2 - 6y - 16 = 0

step 1: Add 16 over to the right side
            y^2 - 6y - 16 = 0 
                           +16   +16
                   y^2 - 6y = 16

step 2: Complete the Square by dividing -6 by 2 and then squaring the
            quotient.
     y^2 - 6y + (-6/2)^2 = 16 + (-6/2)^2
                 y^2 - 6y + 9 = 16 + 9
                 y^2 - 6y + 9 = 25

step 3: Change y^2 - 6y + 9 into the binomial expression (ax+b)^n where a,
             b and n all represent integers

                         (y-3)^2 = 25

step 4: Square root both sides, remember when the right number is square
            to include +/- in front of it.
                        _______           ___
                      V (y-3)^2 = +/- V 25
                              y - 3 = +/- 5
                                   y = 3 +/- 5
                                     ________
                                   | y = 8 , -2 |
                                   |________|

2. The best way to approach this problem is to consider the numbers
    between 0 and 100 as pairs.
   
    For instance 0 + 100 = 100 , 1 + 99 = 100 , 2 + 98 = 100, etc...
    All the way to 49 + 51 being the last pair and 50 stands alone.
    This looks like:
                           (0 + 100) + (1 + 99) + (2 + 98) + ... + (49 + 51) + 50
    There are in total 50 pairs that add up to 100 and then that lonesome 50     which is similar to saying:
                                                        ____
                              50(100) + 50 = |5050|
                                                       |____|

* The famous mathematician Carl Gauss is credited when finding the solution
   to this problem in elementary school in the 1700s, when assigned this for
   busy work by the teacher. He completed the problem incredibly quickly too!

   As such the formula now exist for the sum of any consecutive integers
   which is:
                  S = [(n)(n+1)]/ 2
    where S represents the sum and
               n represents the number of consecutive integers.                          *
Note: This answer could also be derived from disregarding 0 since it adds no
          value to the sum and pairing the numbers as such:
         
          (1 + 100) + (2 + 99) + (3 + 98) + .... + (50 + 51)
         
          there will still be 50 pairs of 101 which multiplies to 5050!

3. I will use f'(x) to represent the derivative... d/dx also works!

    f(x) = 3x * ln(2x^3)
step 1: use CHAIN RULE to take derivative since it is two functions.
            Note: ln(x)'s derivative is 1/x
            Note: derivative power rule applies to 2x^3 function inside of ln 

    f'(x) = 3 * ln(2x^3)+ 3x * (6x^2)/(2x^3)

step 2: Simplify

    f'(x) = 3 * ln(2x^3) + 3x * (3/x)
                ____________
    f'(x) = |3 * ln(2x^3) + 9| , since x's cancel
              |_____________|


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