  # Week Three Written Assignment Math 126

by KIKI
(Chicago,IL USA)

Week Three Written Assignment

Following completion of your weekly readings, complete the exercises in the “Projects” section on page 331 of Mathematics in Our World.

You should be concise in your reasoning. For Project #1, work only equations (a) and (c), but complete all 6 steps (a-f) as shown in the example.

For Project #2, please select at least five numbers; 0 (zero), two even, and two odd. Make sure you organize your paper into separate projects.

The assignment must include (a) all math work required to answer the problems as well as (b) introduction and conclusion paragraphs.

### Comments for Week Three Written Assignment Math 126

 Sep 29, 2011 Solve Quadratic Equations Using Indian Method by: Staff ------------------------------------------------ Part IIIEquation (C) : x² + 12x - 64 = 0 (a) Move the constant term to the right side of the equation. x² + 12x - 64 = 0 x² + 12x - 64 + 64 = 0 + 64 x² + 12x + 0 = 0 + 64 x² + 12x = 0 + 64 x² + 12x = 64 (b) Multiply each term in the equation by four times the coefficient of the x squared term. The coefficient of the x² term is 1. x² + 12x = 64 (4 * 1) * (x² + 12x = 64) (4) * (x² + 12x = 64) (4)*x² + (4)*(12x) = (4)*(64) 4x² + 48x = 256 (c) Square the coefficient of the original x term and add it to both sides of the equation. The coefficient of the original x term is 12. (12)² = 144 4x² + 48x = 256 4x² + 48x + 144 = 256 + 144 4x² + 48x + 144 = 400 (d) Take the square root of both sides. 4x² + 48x + 144 = 400 Sqrt(4x² + 48x + 144) = Sqrt(400) Sqrt(2x + 12)² = Sqrt(20²) Sqrt(2x + 12)² = ±20 2x + 12 = ±20 (to simplify both sides of the equation, divide each side of the equation by 2) (2x + 12)/2 = ±20/2 x + 6 = ±10 (e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x. x + 6 = 10 x + 6 - 6 = 10 - 6 x + 0 = 4 x = 4 (f) Set the left side of equation equal to the negative square root of the number on the right side of the equation and solve for x. x + 6 = -10 x + 6 - 6 = -10 - 6 x + 0 = -16 x = -16 the final solution is: x ∈ {-16, 4} check the solution by substituting the two numerical values of x into the original equation for x = -16 x² + 12x - 64 = 0 (-16)² + 12(-16) - 64 = 0 256 - 192 - 64 = 0 256 - 256 = 0, OK for x = 4 x² + 12x - 64 = 0 (4)² + 12(4) - 64 = 0 16 + 48 - 64 = 0 64 - 64 = 0 0 = 0, OK Project 2: Formula that yields prime numbers. x² - x + 41select at least five numbers; 0 (zero), two even, and two odd.x values selected: 0, 6, 8, 11, 13if x = 0 x² - x + 41= 0² - 0 + 41= 4141 is a prime numberif x = 6 x² - x + 41= 6² - 6 + 41= 7141 is a prime numberif x = 8 x² - x + 41= 8² - 8 + 41= 9797 is a prime numberif x = 11 x² - x + 41= 11² - 11 + 41= 151151 is a prime numberif x = 13 x² - x + 41= 13² - 13 + 41= 197197 is a prime numberHowever the expression x² - x + 41 does yield composite numbers as well:If x = 45 x² - x + 41= 45² - 45 + 41= 20212021 is a COMPOSIT number. Its factors are: 43 and 472021 can be factored into at least two smaller positive integers which not equal to 1.Thanks for writing.Staff www.solving-math-problems.com

 Sep 29, 2011 Solve Quadratic Equations Using Indian Method by: Staff ------------------------------------------------ Part II Project 1: Equation (A) : x² - 2x - 13 = 0 (a) Move the constant term to the right side of the equation. x² - 2x - 13 = 0 x² - 2x - 13 + 13 = 0 + 13 x² - 2x + 0 = 0 + 13 x² - 2x = 0 + 13 x² - 2x = 13 (b) Multiply each term in the equation by four times the coefficient of the x squared term. The coefficient of the x² term is 1. x² - 2x = 13 (4 * 1) * (x² - 2x = 13) (4) * (x² - 2x = 13) (4)*x² + (4)*(- 2x) = (4)*(13) 4x² - 8x = 52 (c) Square the coefficient of the original x term and add it to both sides of the equation. The coefficient of the original x term is -2. (-2)² = 4 4x² - 8x = 52 4x² - 8x + 4 = 52 + 4 4x² - 8x + 4 = 56 (d) Take the square root of both sides. 4x² - 8x + 4 = 56 Sqrt(4x² - 8x + 4) = Sqrt(56) Sqrt(2x - 2)² = Sqrt(2²*2*7) Sqrt(2x - 2)² = Sqrt(2²)*Sqrt(14) (2x - 2) = 2*Sqrt(14) (to eliminate the 2’s on both sides of the equation, divide each side of the equation by 2) (2x - 2)/2 = (2/2)*Sqrt(14) (x - 1) = (1)*Sqrt(14) x - 1 = (1)*Sqrt(14) (e) Set the left side of the equation equal to the positive square root of the number on the right side and solve for x. x - 1 = +Sqrt(14) x - 1 + 1 = Sqrt(14) + 1 x + 0 = Sqrt(14) + 1 x = 1 + Sqrt(14) x = 1 + 3.74166 x = 4.74166 (f) Set the left side of equation equal to the negative square root of the number on the right side of the equation and solve for x. x - 1 = -Sqrt(14) x - 1 + 1 = -Sqrt(14) + 1 x + 0 = -Sqrt(14) + 1 x = 1 - Sqrt(14) x = 1 - 3.74166 x = -2.74166 the final solution is: x ∈ {-2.74166, 4.74166} check the solution by substituting the two numerical values of x into the original equation for x = -2.74166 x² - 2x - 13 = 0 (-2.74166)² - 2(-2.74166) - 13 = 0 7.51669 + 5.48332 - 13 = 0 13 - 13 = 0, OK for x = 4.74166 x² - 2x - 13 = 0 (4.74166)² - 2(4.74166) - 13 = 0 22.4833 - 9.48332 - 13 = 0 13 - 13 = 0, OK ------------------------------------------------