LEARNING TASK 2: Solve the following QE. 1) x² + 12x = 0 2) 3x² + 15x = 0 3) x²-121 = 0 4) 3x² - 12 = 0 5) x²-12x-45 = 0 6) x² + 14x + 49 = 0 7) 2x² -7x-15= 0 8) 2x² - 11x = 21 No Notes.
Answer (1)
Answer:Here are the solutions to the quadratic equations (QE): x² + 12x = 0: Factoring out x, we get x(x + 12) = 0. Therefore, x = 0 or x = -12.3x² + 15x = 0: Factoring out 3x, we get 3x(x + 5) = 0. Therefore, x = 0 or x = -5.x² - 121 = 0: This is a difference of squares, so (x - 11)(x + 11) = 0. Therefore, x = 11 or x = -11.3x² - 12 = 0: Factoring out 3, we get 3(x² - 4) = 0. Then, factoring the difference of squares, 3(x - 2)(x + 2) = 0. Therefore, x = 2 or x = -2.x² - 12x - 45 = 0: Factoring, we get (x - 15)(x + 3) = 0. Therefore, x = 15 or x = -3.x² + 14x + 49 = 0: This is a perfect square trinomial, so (x + 7)² = 0. Therefore, x = -7.2x² - 7x - 15 = 0: Using the quadratic formula or factoring, we get (2x + 3)(x - 5) = 0. Therefore, x = -3/2 or x = 5.2x² - 11x = 21: First, rewrite as 2x² - 11x - 21 = 0. Using the quadratic formula or factoring, we get (2x + 3)(x - 7) = 0. Therefore, x = -3/2 or x = 7.
