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)

1. 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