Math Algebra Quadratic Solutions

UNIT 1 - Solving Equations Solve the equation. (Non-Graphing Calculator) x^3 + 4x^2 - x - 4 = 0 (GROUPING) x^2(x+4) - 1(x+4) = 0 (x+4)(x^2 - 1) = 0 (x+4)(x+1)(x-1) = 0 x = -4, ±1 x^2 + 7x + 4 = 0 (QUADRATIC FORMULA) x = (-7 ± √(7^2 - 414)) / (2*1) x = (7 ± √33) / 2 x^3 + 3x^2 - 4 = 0 (RATIONAL ROOT THEOREM) x = 1 (multi. 2) 1 1 3 -4 1 1 4 0 x^2 + 4x + 4 = 0 (x+2)^2 = 0 x = -2 10x^3 - 25x^2 - 15x = 0 (GCF) 5x(2x^2 - 5x - 3) = 0 5x(2x+1)(x-3) = 0 x = 0, -1/2, 3 4x^2 + 8x - 5 = 0 (FACTOR) (2x-1)(2x+5) = 0 x = 1/2, -5/2 x^3 + 4x^2 + 3x - 2 = 0 (RATIONAL ROOT THEOREM) ±1, ±2 1 1 4 3 -2 1 1 5 8 x^2 + 2x - 1 = 0 x = (-2 ± √(2^2 - 41(-1))) / (2*1) x = (-2 ± √8) / 2 x = -2, -1 ± √2 2(x+1)^2 + 5(x+1) - 3 = 0 (U SUBSTITUTION) let u = x+1 2u^2 + 5u - 3 = 0 (2u-1)(u+3) = 0 u = 1/2, -3 x+1 = 1/2, -3 x = -1/2, -4 x^2 + 2x + 2 = 0 (COMPLETE THE SQUARE OR QUADRATIC FORMULA) x^2 + 2x = -2 x^2 + 2x + 1 = -2 + 1 (x+1)^2 = -1 x+1 = ±i x = -1 ± i 9. x^4 - 6x^3 + 4x^2 + 15x + 4 = 0 (RATIONAL ROOT THEOREM) ±1, ±2, ±4 1 -6 4 15 4 4 -2 2 16 1 -2 2 16 0 3x^2 + 2x - 16 = 0 x = (-2 ± √(2^2 - 4*3*(-16))) / (2*3) x = (-2 ± √196) / 6 x = (-2 ± 14) / 6 x = -1, 4, (3 ± √13) / 2 10. 4x^2 + 5x = x^2 + 32 (FACTORING) 4x^2 + 20x = x^2 + 32 3x^2 + 20x - 32 = 0 (3x - 4)(x + 8) = 0 x = 4/3, -8 11. 3x^3 - x^2 + 6x - 2 = 0 (GROUPING) x^2(3x-1) + 2(3x-1) = 0 (3x-1)(x^2 + 2) = 0 3x - 1 = 0 x = 1/3 x^2 + 2 = 0 x = ±i√2 12. x^4 - 13x^2 + 40 = 0 (QUADRATIC FORM) (x^2 - 8)(x^2 - 5) = 0 x^2 - 8 = 0 x^2 = 8 x = ±√8 = ±2√2 x^2 - 5 = 0 x^2 = 5 x = ±√5 13. x^4 - 4x^3 + 10x^2 - 24x + 24 = 0 (RATIONAL ROOT THEOREM) ±1, ±2, ±3, ±4, ±6, ±8, ±12, ±24 1 -4 10 -24 24 2 -2 8 -16 8 2 -6 2 -12 0 x^2 - 6x + 12 = 0 x = (6 ± √(-6)^2 - 4*1*12) / (2*1) x = (6 ± √-24) / 2 x = 3 ± i√6 14. Suppose a bullet was fired into the air with an initial velocity of 848 ft/sec. Its height after t seconds is given by h(t) = -16t^2 + 848t. (CALCULATOR) a) When will the bullet reach 10,000 ft? 10000 = -16t^2 + 848t 0 = t^2 - 53t + 625 t = (53 ± √(53^2 - 41625)) / (2*1) t = (53 ± √2809) / 2 t = (53 ± 17.6) / 2 t = 35.3, 17.7 seconds b) When will the bullet reach 12,000 ft? 12000 = -16t^2 + 848t 0 = t^2 - 53t + 750 t = (53 ± √(53^2 - 41750)) / (2*1) t = (53 ± √-191) / 2 The bullet won't reach 12,000 ft. c) When will the bullet hit the ground? 0 = -16t^2 + 848t 0 = -16t(t - 53) t = 0, 53 seconds 15. A ball is thrown across a baseball field. Its path is modeled by the equation h = -0.005x^2 + x + 6, where x is the distance the ball traveled horizontally and h is the height of the ball. Both x and h are measured in feet. (CALCULATOR)   a) What is the maximum height attained by the ball? h = -0.005(100)^2 + 100 + 6 = 56 ft   b) How many feet horizontally does the ball travel when it reaches its maximum? 100 ft 16. A farmer has 600 ft of fencing to fence in a pen for his goats. He plans to create a rectangular pen along the side of his barn and will not need to install the fence along the barn. What is the maximum area he can enclose? (CALCULATOR)   A(x) = x(600 - 2x) A(x) = -2x^2 + 600x x = -600 / (2(-2)) = 150 A = 150(600 - 2(150)) = 45,000 ft^2 17. Find a 3rd-degree polynomial with zeros 2 and -5 and with a leading coefficient of 4. (x-2)^2(x+5) (x^2 - 4x + 4)(x+5) x^3 + 5x^2 - 4x^2 - 20x + 4x + 20 x^3 + x^2 - 16x + 20 4x^3 + 4x^2 - 64x + 80 OR (x-2)(x+5)^2 (x-2)(x^2 + 10x + 25) x^3 + 10x^2 + 25x - 2x^2 - 20x - 50 x^3 + 8x^2 + 5x - 50 4x^3 + 32x^2 + 20x - 200 18. Find a polynomial that has zeros 6 and 1-i. (CALCULATOR) (x-6)(x-(1-i))(x-(1+i)) (x-6)[(x-1)^2 - i^2] (x-6)(x^2 - 2x + 1 + 1) (x-6)(x^2 - 2x + 2) x^3 - 8x^2 + 14x - 12