Quadratic Equations

Quadratic Equations I am asked to complete the exercises in the rolls part on page 397 of Mathematics in Our World. I am to be concise in my reasoning. P choose see conquer the stairs my work for calculate iodin and Project Two. Project One: Equation (C): x² + 10x - 40 = 0 (a) snuff it the eternal experimental condition to the sound perspective of the par. X² + 10x - 40 = 0 x² + 10x - 40 + 40 = 0 + 40 x² + 10x + 0 = 0 + 40 x² + 10x = 0 + 40 x² + 10x = 40 (b) Multiply to each one term in the par by four measure the coefficient of the x unanimousd term. The co-efficient of the x² term catchs 1. x² + 10x = 40 (4 * 1) * (x² + 10x = 40) (4) * (x² + 10x = 40) (4)*x² + (4)*(10x) = (4)*(40) 4x² + 40x = clx (c) Square the coefficient of the original x term and minimal brain dysfunction it to both posts of the equation. The co-efficient of the original x term is 10. (10)² = 100 4x² + 40x = 160 4x² + 40x + 100 = 16 0+ 100 4x² + 40x + 160 = 260 (d) Take the self-colored root of both sides. 4x² + 40x + 100 = 260 Sqrt(4x² + 40x + 160) = Sqrt(260) Sqrt(2x + 10)² = Sqrt(10²) Sqrt(2x + 10)² = ±10 2x + 10 = ±10 (to modify both sides of the equation, divide each side of the equation by 2) (2x + 10)/2 = ±10/2 x + 6 = ±10 (e) Set the left side of the equation equal to the dogmatic square root of the tot 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} Project Two Formula that yields prime numbers: x² - x + 41 convey at least five dollar bill numbers; 0 (zero), devil even, and two odd. x determine selected: 0, 7, 9, 12, 14 if x = 0 x² - x + 41 = 0² - 0 + 41 = 41 if x = 7 x² - x + 41 = 7² - 7 + 41 = 83 if x = 9 x² - x + 41 = 9² - 9 + 41 = 113 if x = 12 x² - x + 41 = 12² - 12 + 41 = 173 if x = 14 x² - x + 41 =...If you want to lease a full essay, order it on our website: BestEssayCheap.com

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