MATHEMATICS MAGAZINE Home Math Tests Math Tutors Contact Us CORRESPONDENTS A NEW METHOD TO SOLVE QUADRATIC EQUATIONS AND TO FACTOR TRINOMIALS   By Thomas Nguyen (Copyrights TX 6 – 493 – 074)   Abstract:    A new method presented in this paper can be considered as a special method to solve a quadratic equation (First, it can solve any quadratic equations. Second, solutions can be found very fast and so simple in many occasions)   For the problem of solving the quadratic equation, the advantage of this new method is in this case: If we can guess two numbers n1 & n2 which satisfy the following condition (Sum & Product):   n1 + n2 = b    and   n1*n2 = a*c                                                (*)   Then the solution of the quadratic equation can be written out right away: (1)   Summarize the old method and the new method for solving a quadratic equation:  For example, solve equation:  6x2 +5x + 1 = 0. We have:  n1*n2 = a*c=6 and n1 + n2 = b = 5. We can guess n1=2 & n2=3. When we know n1 & n2, the problem can be considered as “done”. By applying Formula (1), we have:   x = -2/6=-1/3 and x = -3/6=-1/2. It’s clear that the new method is fater and simpler than the old method in this case..      “What happens in the case we cannot guess n1 and n2?”, In that case, we just follow the following steps:     - Step 1: Use the following formula to find n1 & n2 first: (2)   Step2: Use (1) to write out the solutions. In this case, these two methods are equivalent in time to find solutions.   For the problem of factoring a trinomial, we can use the following formula to write out the answer, where n1 & n2 come from guessing or computing as above (3)   This formula saves time for us. We don’t need to split the midterm (bx) into 2 parts:  n1x & n2x. We don’t need to do “factor by grouping”. We don’t need to do crossing diamond diagram, etc.                                                1. Examples:                                                      a. Solve the equation:           2x2 + 11x + 15 = 0                     Since  a*c = 2*15 = 30 and  b = 11, we can guess  n1 = 5  and   n2 = 6                     Applying Formula (1), we have:      x = - 5/2   and     x = - 6/2 = -3                   b. Solve the equation: Sine  a*c = (1/5)*(-10) = -2    and   b = -1 , we can guess     n1 = - 2     and     n2 =  1            Applying Formula (1), we have:   x = -  (-2) ÷ (1/5) = 10 ; x =  - 1 ÷ (1/5) =  - 5                     c. Solve the equation:        -3x2 – x + 2 = 0      (assume that you cannot guess)                     We can compute n1 & n2 (using Formula (2)) Using Formula (1), we have:      x = - (-3)/ (-3) = -1   and    x = -2/(-3) = 2/3                   d. Factor:      5x2 + 11x + 2                     Since  a*c = 10 and b = 11, we can guess   n1 = 1 and  n2 = 10                     Applying Formula (3), we have:    5x2 + 11x + 2 = 5 (x + 1/5)(x + 10/5) = (5x + 1)(x + 2)       2. Proof: (Abstract)   Let x1 & x2 be solutions of a quadratic equation. According to the old method: Where It’s easy to verify:                 x1* x2 = c/a         and        x1 + x2 = -b/a         Multiply both sides of two above equations by ( a2) ; we have: Or                                    (-a*x1)(-a*x2) = c*a     and    (-a*x1) + (-a*x2) = b                                                    (4)   From (4) we can realize that the two numbers, which satisfied condition (*), are:                                               n1 = (- a*x1)     and       n2 = (- a*x2)                                                                    (5)   Therefore, we can have the solutions of the quadratic equation (Formula (1)): Solve for n1 and use condition:  n1 + n2,= b, we will have Formula (2) If we replace x1 & x2 into We will have Formula (3): (The set of Formula (1), Formula (2), and Formula (3) was named “3JCN Formula” after my children names.)   San Diego, Spring 2004