Solving quadratic equation using AC method (Algebra 2)

Let's continue with our series on solving quadratic equations.   If you can think of what a quadratic equation means, you will know that the quadratic equation is a graph of a parabola.  Unless the roots (x intercepts) of the quadratic equation are an imaginary pair of solution, typically the solution will have 1-2 real roots.  Having 1-2 real roots also means that graphically these roots represent the x intercept(s) of the parabola.  In this article, I will show you how you can solve for these roots using the AC method.

NOTE:  When you are given a particular quadratic equation, you can use various methods to solve for the final solution.  While it is true that the quadratic formula will always give the solution to the quadratic equation, some times it might be easier to deploy other methods to solve the same problem.

In our example, we will try to solve the quadratic equation of the following:

2x^2 = -11x - 5

In order to use the AC method, you have to make the equation into its standard form first.

Ax^2 + Bx + C = 0

Which makes our equation to be:

2x^2 + 11x + 5 = 0

The AC method of factoring is basically a method to split the middle term bx into 2 separate terms so that you can eventually factor the trinomial using grouping.  In order to split the middle term (in this case 11x), we will need to find the factors that make up the product of the coefficient A and C. 

The coefficients in this case are:

A = 2, B = 11, c = 5

So the product of AC = 2 x 5 = 10

We will then need to find factors of AxC that will add up to the coefficient B.  In other word, we will need to find the factors of 10 that will add up to 11.

Factors of 10:

1. 1 x 10 => 1 + 10 = 11
2. -1 x -10 => -1 + -10 = -11
3. 2 x 5 => 2 + 5 = 7
4. -2 x -5 => -2 + -5 = -7

1. 5x^2 - 13x + 6 = 0
2. 3x^2 - 17x - 6 = 0
3. 6x^2 + 7x + 2 = 0