Mat126 Week 3

This week’s assignment involves a method of solving a quadratic equation that purports to have originated from India. I shall use this methodology to complete (a) and (c), as assigned. I shall also complete an assignment involving a previously derived formula that may yield prime numbers . I shall conclude the assignment with a summary of what I have learned in completing the two assigned projects.

Project 1:

(a) Solve: x2 – 2x – 13 = 0 using the 6 step method described on page 331 of Mathematics in Our World.

1.) “Move the constant term to the right side of the equation.

x2 – 2x – 13 = 0 x2 – 2x = 13

2.) “Multiply each term in the equation by four times the coefficient of the x2 term.”:

x2 – 2x = 13
4x2 – 8x = 52

3.) “Square the coefficient of the original x term and add it to both sides of the equation.”:

4x2 – 8x = 52
4x2 – 8x + 4 = 52 + 4
4x2 – 8x + 4 = 56

4.) “Take the square root of both sides.”:

4x2 – 8x + 4 = 56
2x – 2 = +/- √56

5.) “Set the left side of the equation equal to the positive square root of the number on the right side and solve for x.”:

2x – 2 = √56
2x = √56 +2 x = (√56 +2)/2 x = 4.7416573867739414

6.) “Set the left side of the equation equal to the negative square root of the number on the right side of the equation and solve for x.”:

2x – 2 = -√56
2x = -√56 + 2 x = -(√56 +2)/2 x = -2.7416573867739414 For Project 1: (a), x = 4. 7416573867739414 and x = -2.7416573867739414.

Proof: x2 – 2x – 13 = 0

(4. 7416573867739414)2 – 2(4. 7416573867739414) – 13 = 0
(22.483314773547882) – (9.483314773547882) – 13 = 0

(-2.7416573867739414)2 – 2(-2.7416573867739414) – 13 = 0
(7.516685226452117) + (5.483314773547883) – 13 = 0

(c) Solve: x2 + 12x – 64 = 0 using the 6 step method described on page 331 of Mathematics in Our World.

1.) “Move the constant term to the right side of the equation.

x2 + 12x – 64 = 0 x2 + 12x = 64

2.) “Multiply each term in the equation