Week 3 Assignment
May 20, 2012
For this assignment I will be working 2 projects. For project #1 I will solve equations ( a) x² - 2x - 13 = 0 and ( c ) x² + 12x - 64 = 0 using steps a-f from page 379 Mathematics in Our World. For project #2, I selected five numbers consisting of 0, two even, and two odd. The projects actually comes from an interesting method for solving quadratic equations. The methods came from India (Bluman, 2011.) I will show my work by starting with Project 1 (A) and (C) and then move to Project 2. For Project 1 (A) and (C), step (A) requires me to move the constant term to the right side of the equation. (B) asks me to multiply each term in the equation by four times the coefficient of the x squared term. (C) requires me to square the coefficient of the original x term and add it to both sides of the equation. For (D) I must take the square root of both sides. (E) wants me to set the left side of the equation equal to the positive square root of the number on the right side and solve for x. Lastly, for (F) I will 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. Below I will show you the work for each step for Project 1 (A) The I will do the same for (C). Step (A) x² - 2x - 13 = 0 wants me to move the constant term to the right side of the equation. To do this I wrote x² - 2x - 13 + 13 = 0 + 13 which I got me to the answer x² - 2x = 13. Step (B) asks me to multiply each term in the equation by four times the coefficient of the x squared term. From the answer from step (A) x² - 2x = 13 using 1 as the coefficient term and multiplying it by four the answer is: 4x² - 8x = 52 Below is the work I did to get the this answer (4 * 1) * (x² - 2x = 13)
(4) * (x² - 2x = 13)
(4)*x² + (4)*(- 2x) = (4)*(13)
4x² - 8x = 52
Step(C) I will square the coefficient of the original x term and add it to...
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