The an analytical proof of the quadratic formulas used to solve quadratic equations is presented. Examples on how to use the quadratic formulas and the discriminant to solve various questions related to quadratic equation are also presented with detailed explanations.
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Analytical Proof of the Quadratic FormulasA quadratic equation in the standard form is given bya x2 + b x + c = 0 where a, b and c are constants with a not equal to zero. Solve the above equation to find the quadratic formulas
Questions on Quadratic EquationsQuestion 1Solve the equationSolution to Question 1We first identify the coefficients a, b and c.a = 2 , b = - 7 and c = 4 The discriminant ? = b2 - 4 a c is now calculated. ? = b2 - 4 a c = (-7)2 - 4 (2)(4) = 17 The discriminant is positive and therefore the equation has two real solutionsUse given by the quadratic formulas: x1 = (- b + √( b2 - 4 a c)) / (2 a) = (7 + √(17)) / 4 = 7 / 4 + √(17) / 4 x2 = (- b - √ ( b2 - 4 a c)) / (2 a) = (7 - √(17)) / 4 = 7 / 4 - √(17) / 4 Question 2Solve the equationSolution to Question 2We first identify the coefficients a, b and c.a = 1 , b = - 2 and c = 5 The discriminant ? = b2 - 4 a c is now calculated. ? = b2 - 4 a c = (-2)2 - 4 (1)(5) = - 16 The discriminant is negative and therefore the equation has two complex solutions. In complex numbers, we define the imaginary unit i as i = √(-1) The Solutions to the equation are given by x1 = (- b + √( b2 - 4 a c)) / (2 a) = (2 + √(-16)) / 2 x2 = (- b - √ ( b2 - 4 a c)) / (2 a) = (2 - √(-16)) / 2 We now simplify √(-16) as follows √(-16) = √(-1)√(16) = 4 i , where i is the imaginary unit defined above. We now simplify the two solutions x1 = (2 + √(-16)) / 2 = (2 + 4 i) / 2 = 1 + 2 i x2 = (2 - √(-16)) / 2 = (2 - 4 i) / 2 = 1 - 2 i Question 3How many solutions does the equationSolution to Question 3We first write the given equation in standard form (with the zero on the right) in the form a x2 + b x + c = 0.2x 2 + 3x + 9 = 0 We now identify the coefficients a, b and c. a = 2 , b = 3 and c = 9 The sign of the discriminant ? = b2 - 4 a c gives the number of solutions to the equation. ? = b2 - 4 a c = (3)2 - 4 (2)(9) = - 63 The discriminant ? is negative and therefore the equation has two complex solutions. Question 4Find all values for the parameter m so that the equationSolution to Question 4We first identify the coefficients a, b and c.a = 1 , b = m and c = 4 The sign of the discriminant ? = b2 - 4 a c gives the number of solutions to the equation. ? = b2 - 4 a c = m 2 - 4 (1)(4) = m2 - 16 For the equation to have two real solutions, the discriminant ? must be positive. Hence we need to solve the following inequality in m m2 - 16 > 0 Factor the right side of the inequality (m - 4)(m + 4) > 0 The values of m for which the given equation has two real solutions are the solutions to the above inequality and are given by the interval ( - ? , - 4) ? (4 , ?) Question 5Find all values for the parameter k so that the equationSolution to Question 5We first write the equation in standard form.- x 2 + (k + 1) x - 2 - k = 0 We now identify the coefficients a, b and c. a = - 1 , b = k + 1 and c = - 2 - k The discriminant ? = b2 - 4 a c is given by ? = b2 - 4 a c = (k + 1)2 - 4 ( - 1)( - 2 - k) = (k + 1)2 - 8 - 4 k For the equation to have one solution, the discriminant ? must be equal to zero. Hence we need to solve the following equation in k (k + 1)2 - 8 - 4 k = 0 Expand the above equation, simplify and rewrite it as k2 - 2 k - 7 = 0 Solve the above quadratic equation in k using the quadratic formulas to obtain the two values of k for which the given equation has one solution only. ? = (-2)2 - 4 (1)(-7) = 32 k1 = ( 2 + √32) / (2 * 1) k2 = ( 2 - √32) / (2 * 1) which simplifies to k1 = 1 + 2√2 and k2 = 1 - 2√2 NOTE: As an exercise, use one value of k found above in the given equation and solve it; you must obtain one solution only.
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