use the discriminant to determine the number of solutions of the quadratic equation, and whether the solutions are real or complex. Note: It is not necessary to find the roots; just determine the number and types of solutions. ---------------------------------- Just learn these rules: Ax² + Bx + C =0 (or any letter besides x) has the discriminant B² - 4AC. Calculate it. 1. If the discriminant is 0, there is one real solution. 2. If the discriminant is negative, there are two conjugate complex solutions. 3. If the discriminant is positive, there are two real solutions. A. If the discriminant is a perfect square, the two real solutions are rational. B. If the discriminant is not a perfect square, the two real solutions are irrational. ---------------- 2x² + x - 1 = 0 Give the x a coefficient of 1: 2x² + 1x - 1 = 0 Compare that to Ax² + Bx + C = 0 A = 2, B = 1, and C = -1 Calculate discriminant B² - 4AC (1)² - 4(2)(-1) 1 + 8 9 9 is positive, so there by rule 3 above there are two real solutions. 9 is also a perfect square, so by rule 3A the two real solutions are rational. ------------------ 4/3x² - 2x + 3/4 = 0 Compare that to Ax² + Bx + C = 0 A = 4/3, B = -2, and C = 3/4 Calculate discriminant B² - 4AC (-2)² - 4(4/3)(3/4) -4 -4 is negative, so there by rule 2 above there are two conjugate complex solutions. ----------------------- 2x² + 5x + 5 = 0 Compare that to Ax² + Bx + C = 0 A = 2, B = 5, and C = 5 Calculate discriminant B² - 4AC (5)² - 4(2)(5) 25 - 40 -13 -13 is negative, so there by rule 2 above there are two conjugate complex solutions. ------------------------ 3z² + z - 1 = 0 Give the z a coefficient of 1 3z² + 1z - 1 = 0 Compare that to Ax² + Bx + C = 0 A = 3, B = 1, and C = -1 Calculate discriminant B² - 4AC (1)² - 4(3)(-1) 1 + 12 13 13 is positive, so there by rule 3 above there are two real solutions. 13 is not a perfect square, so by rule 3B the two real solutions are irrational. ----------------------------------- m² + m + 1 = 0 Give the m² and the m coeficients of 1 each 1m² + 1m + 1 = 0 Compare that to Ax² + Bx + C = 0 A = 1, B = 1, and C = 1 Calculate discriminant B² - 4AC (1)² - 4(1)(1) 1 - 4 -3 -3 is negative, so there by rule 2 above there are two conjugate complex solutions. Edwin