# Mat126 Week 3

Topics: Polynomial, Prime number, Real number Pages: 5 (671 words) Published: January 28, 2013
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 by four times the coefficient of the x2 term.”:

x2 + 12x = 64
4x2 + 48x = 256

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

4x2 + 48x = 256
4x2 + 48x + 144 = 256 + 144
4x2 + 48x + 144 = 400

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

4x2 + 48x + 144 = 400
2x + 12 = +/- 20

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 + 12 = 20
2x = 20 - 12
x = 8/2
x = 4

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 + 12 = -20
2x = -20 - 12
x = -32/2
x = -16

For Project 1: (c), x = 4 and x = -16.

Proof:
x2 + 12x – 64 = 0

(4)2 + 12(4) – 64 = 0
(16) + (48) – 64 = 0

(-16)2 + 12(-16) – 64 = 0
(256) + (-192) – 64 = 0

Project 2:

The formula “x2 - x + 41” is a formula that yields prime numbers, according to page 331 of Mathematics in Our World. The following work uses different values as x in an attempt to yield a prime number. We shall use the following numbers to see if this statement is true: 0, 12, 68, 39, & 115.

a.)x2 - x + 41 (x = 0)
(0)2 – (0) + 41 =
0 - 0 + 41 = 41

b.)x2 - x + 41 (x = 12)
(12)2 - (12) + 41 =
144 - 12 + 41 = 197

c.)x2 - x + 41 (x = 68)
(68)2 – (68) + 41 =
4,624 - 68 + 41 = 4,597

d.)x2 - x + 41 (x = 39)
(39)2 – (39) + 41 =
1,521 - 39 + 41 = 1,523

e.)x2 - x + 41 (x = 115)
(115)2 – (115) + 41 =
13,225 - 115 + 41 = 13,151

Since all 5 results of x conclude with prime numbers (41, 197, 4597, & 13151), this statement appears to be accurate, x2 - x + 41 yields prime numbers.