**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):
* *
*n*_{1} + n_{2} = b and n_{1}*n_{2} = 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: 6x*^{2} +5x + 1 = 0. We have: n_{1}*n_{2} = a*c=6 and n_{1} + n_{2} = b = 5. We can guess n_{1}=2 & n_{2}=3. When we know n_{1} & n_{2}, 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 n___{1} and n_{2}?”, 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: n*_{1}x & n_{2}x. 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: 2x^{2} + 11x + 15 = 0
Since a*c = 2*15 = 30 and b = 11, we can guess n_{1} = 5 and n_{2} = 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: -3x^{2} – x + 2 = 0 (assume that you cannot guess)
We can compute n_{1} & n_{2} (using Formula (2))
Using Formula (1), we have: x = - (-3)/ (-3) = -1 and x = -2/(-3) = 2/3
d. Factor: 5x^{2} + 11x + 2
Since a*c = 10 and b = 11, we can guess n_{1} = 1 and n_{2} = 10
Applying Formula (3), we have: 5x^{2} + 11x + 2 = 5 (x + 1/5)(x + 10/5) = (5x + 1)(x + 2)
2. __Proof: (Abstract)__
Let x_{1} & x_{2} be solutions of a quadratic equation. According to the old method:
Where
It’s easy to verify: x_{1}* x_{2} = c/a and x_{1} + x_{2} = -b/a
Multiply both sides of two above equations by ( a^{2}) ; we have:
Or
(-a*x_{1})(-a*x_{2}) = c*a and (-a*x_{1}) + (-a*x_{2}) = b (4)
From (4) we can realize that the two numbers, which satisfied condition (*), are:
n_{1} = (- a*x_{1}) and n_{2} = (- a*x_{2}) (5)
Therefore, we can have the solutions of the quadratic equation (Formula (1)):
Solve for n_{1} and use condition: n_{1} + n_{2},= b, we will have Formula (2)
If we replace x_{1 }& x_{2} 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 |