# How to compare the AC Method (factoring by grouping) and the new Transforming Method in solving quadratic equations?

May 28, 2017

Solving quadratic equations by the new Transforming Method

#### Explanation:

A good way to compare these 2 methods is solving a sample of quadratic equation.
The Transforming Method. Solve
$y = 16 {x}^{2} - 62 x + 21 = 0$
Transformed equation:
$y ' = {x}^{2} - 62 x + 336 = 0$ -->( ac = 336)
Proceeding: Find the 2 real roots of the transformed equation y', then, divide them by a = 16.
Find 2 numbers knowing sum (-b = 62) and product (ac = 336).
Compose factor pairs of (336) --> ...(4, 82)(6, 56). This sum is (6 + 56 = 62 = -b). Then, the 2 real roots of y' are: 6 and 56.
Back to y, the 2 real roots are:
$x 1 = \frac{6}{a} = \frac{6}{16} = \frac{3}{8}$, and $x 2 = \frac{56}{16} = \frac{7}{2}$.

May 28, 2017

Solving quadratic equation by the AC Method (splitting the middle term)

#### Explanation:

$y = 16 {x}^{2} - 62 x + 21 - 0$
Proceed to split the middle term by proceeding as follows:
Find 2 numbers knowing sum (b = -62) and product
(ac = 16*21 = 336)
Compose factor pairs of (336):
... (-4, - 82)(-6, -56). This sum is (-62) and its product is (336).
Re-write the equation and split the middle term (-62x) into (-6x) and
(- 56x)
$y = 16 {x}^{2} - 6 x - 56 x + 21$
$y = 2 x \left(8 x - 3\right) - 7 \left(8 x - 3\right)$
$y = \left(8 x - 3\right) \left(2 x - 7\right)$
Solve the 2 binomials:
$\left(8 x - 3\right) = 0$ --> $x = \frac{3}{8}$
$\left(2 x - 7\right) = 0$ --> $x = \frac{7}{2}$

NOTE. The new Transforming Method (Google Search) avoids the lengthy factoring by grouping and solving the 2 binomials.
After you find the 2 numbers (-6) and (-56). Take the opposite of them, (6) and (56), then divide them by a = 16, you immediately get the 2 real roots. You don't need to proceed factoring by grouping further.