Lesson Solving quadratic equations: Compare the factoring "ac method" and the new Diagonal Sum Method .

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SOLVING QUADRATIC EQUATIONS: COMPARE THE FACTORING "ac METHOD" AND THE DIAGONAL SUM METHOD.

A. GENERALITIES. When a given quadratic equation can be factored, there are 2 best methods to solve it: the factoring "ac Method" (You Tube) and the new Diagonal Sum Method.
This lesson, first describes in general these 2 methods, then compares them through working examples.

1. THE "ac factoring method". 
It is also called the "box method" (You Tube.com), with a little variation. The concept of the method is to make the given equation be factored into 2 binomials in x, by replacing the term (bx) by the 2 terms (b1.x) and (b2.x) that satisfy the 2 conditions: 1) the product b1.b2 = ac; 2) the sum (b1 + b2) = b.

Example 1. Solve the equation: 5x^2 + 6x - 8 = 0. 
Solution. Find 2 numbers that the product is (ac = -40) and their sum is b = 6. Proceed:[(-1, 40)(1, -40),(-2, 20)(2, -20),(-4, 10),OK]. Next, substitute the term (6x) by the 2 terms (-4x) and (10x) in the inequality, then put in common factor: 
5x^2 - 4x + 10x - 8 = 5x(x + 2)- 4(x + 2) = (x + 2)(5x - 4) = 0
Then solve the 2 binomials for x: (x + 2) = 0 ----> x = -2 , and (5x - 4) = 0 ----> x = 4/5

2. The new Diagonal Sum Method.
 
Its concept is direct finding the 2 real roots, in the form of 2 fractions, knowing their sum (-b/a) and their product (c/a). Its applies 2 rules. The first rule is the Rule of Signs that shows the signs (- or +) of the 2 real roots before proceeding. The second rule is the Rule of the Diagonal Sum. If (c1/a1, c2/a2) is a pair of real roots, the sum (c1a2 + c2a1) is called the diagonal sum of a root pair, and it must equal to -b. See book titled: "New methods for solving quadratic equations and inequalities" (Amazon e-book 2010). 
This method directly selects the probable root pairs from the (c/a) setup. The numerator of this setup contains all factor pairs of c. The denominator contains all factor pairs of a. Next, it uses mental math to calculate their diagonal sums and find the one that equals to -b.

Example 1. Solve: 5x^2 + 6x - 8 = 0.
Solution. Roots have opposite signs. The constant c has 2 factor pairs: (1, 8) and (2, 4). Select the 4 probable root pairs that come out from the two (c/a) setups: (-1, 8)/(1, 5) and (-2, 4)/(1, 5). The 2 roots having opposite sign, by convention, always put the negative sign (-) in front of the first number of the pair. 
Since -1 is not a real root, there are 3 probable root pairs: (-1/5, 8/1),(-2/1, 4/5),(-2/5, 4/1). The diagonal sum of the second pair is: -10 + 4 = -6 = -b. The 2 real roots are -2 and 4/5.

B. WHEN a = 1 - SOLVING THE QUADRATIC EQUATION: x^2 + bx + c = 0.

a. Example 2. Solve: x^2 - 11x - 102 = 0.

1. The Diagonal Sum method. In the case a = 1, the diagonal sum reduces to the sum of the 2 real roots. Roots have opposite signs. Write factor pairs of c = -102. They are: 
(-1, 102),(-2, 51),(-3, 34),(-6, 17)...This sum is 17 - 6 = 11 = -b. The 2 real roots are -6 and 17.

2. The factoring "ac method". Find 2 numbers that the product is (ac = -102) and the sum is b = -11. Proceed:[ (-1, 102)(1, -102)(-2, 51)(2, -51)(-3, 34)(3, -34)(-6, 17)(6, -17) OK]. Next, replace in the inequality the term (-9x) by the 2 terms (6x) and (-17x):
x^2 + 6x - 17x - 102 = x(x - 17) + 6(x - 17) = (x - 17)(x + 6) = 0.
Next, solve the 2 binomials for x:
(x - 17) = 0 ----> x = 17; (x + 6) = 0 ----> x = -6.

3. Remark. The Diagonal Sum method is simpler and doesn't require factoring. The Rule of Signs helps reduce in half the number of permutations (or test cases). In addition, the new method saves the time used for factoring and for solving the 2 binomials for x.

b. Example 3. Solve: -x^2 + 28x - 96 = 0.

1. The Diagonal Sum method. Both roots are positive. Write factor pairs of ac = 96. They are: (1, 96),(2, 48),(3, 32),(4, 24)...This sum is 4 + 24 = 28 = b. According to the Diagonal Sum Rule, when a is negative, the 2 real roots are 4 and 24.

2. The factoring "ac method". Find 2 numbers with product (ac = 96) and sum (b = 28).
Proceed:[(-1, -96)(1, 96)(-1, -48)(2, 48)(-3, -32)(3, 32)(-4, -24)(4, 24) OK]. Next replace in the inequality the term (28x) by the 2 terms (4x) and (24x):
-x^2 + 4x + 24x - 96 = -x(x - 4) + 24(x - 4) = (x - 4)(24 - x) = 0.
Next, solve the 2 binomials for x:
(x - 4) = 0 ----> x = 4; (24 - x) = 0 ---> x = 24.

C. WHEN a IS NOT 1 - SOLVING THE QUADRATIC EQUATION ax^2 + bx + c = 0.

a. When both a and c are prime numbers.

Example 4. Solve: 7x^2 - 76x - 11 = 0.

1. The Diagonal Sum method. Roots have opposite signs. Since -1 is not a real root, there is unique
probable root pair: (-1/7, 11/1). Its diagonal sum is 77 - 1 = 76 = -b. The 2 real roots are -1/7 and 11.

2. The factoring "ac method". Find 2 numbers; ac = 77 and sum is -76. Proceed:[(-1, 77)(1, -77)OK]. Next, replace in the equation the term (-76x) by the 2 terms (x) and (-77x): 7x^2 + x - 77x - 11 = 7x(x - 11) + (x - 11) = (x - 11)(7x + 1) = 0.
Next, solve the 2 binomials: (x - 11) = 0 ---> x = 11; (7x + 1) = 0 ---> x = -1/7.

b. When a and c are small numbers and may contain themselves one (or 2) factors.

Example 5. Solve: 8x^2 - 22x - 13 = 0.

1. The Diagonal Sum method. Roots have opposite signs. The (c/a) setup: (-1, 13)/(1, 8).(2, 4). Before proceeding, look to simplify the (c/a) setup. Since b = 22 is an even number, we can eliminate the pair(1, 8) from the denominator because it give odd-number diagonal sums. The remainder (c/a) is: (-1, 13)/(2. 4) that leads to 2 probable root pairs:(-1/2, 13/4) and (-1/4, 13/2). The diagonal sum of the first pair is: -4 + 26 = 22 = -b. The 2 real roots are -1/2 and 13/4.

2. The factoring "ac method". Find 2 numbers: ac = -104, and sum = -22. Proceed:[-1, 104)(1, -104)(-2, 52)(2, -52)(-4, 26)(4, -26) OK]. Replace -22x by 4x and -26x.
8x^2 + 4x - 26x - 13 = 4x(2x + 1) - 13(2x + 1) = (2x + 1)(4x - 13) = 0.
(2x + 1) = 0 ---> x = -1/2 ; (4x - 13) = 0 ---> x = 13/4.

Example 6. Solve: 6x^2 + 17x - 14 = 0.

1. Diagonal Sum method. Roots have opposite signs. The (c/a) setup: (-1, 14),(-2, 7)/(1, 6),(2, 3).
Before proceeding, we can eliminate the pairs (-1. 14)/(1, 6) because they will give large diagonal sums (while b = 17). The remainder(c/a):(-2, 7)/2, 3) gives unique probable root pair: (-2/3, 7/2). Its diagonal sum is: 21 - 4 = 17 = b. According to the Diagonal Sum Rule, when the diagonal sum equals to b, the answer is the opposite root pair. The 2 real roots are: 2/3 and -7/2.

2. The factoring "ac method". Find 2 numbers, product ac = -84.; sum = 17. Proceed:[(-1, 84)(1, -84)(-2, 42)(2, -42)(-3, 28)(3, -28)(-4, 21) OK]. 
6x^2 - 4x + 21x - 14 = 2x(3x - 2) + 7(3x - 2) = (3x - 2)(2x + 7) = 0.
(3x - 2) = 0 ---> x = 2/3 ; (2x + 7 = 0 ---> x = -7/2.

c. When a and c are large numbers and may contain themselves many factors. Examples: 12x^2 + 5x - 72 = 0; 45x^2 + 74x + 24 = 0; 24x^2 + 59x + 36 = 0. These cases are considered complicated because there are a lot of permutations involved. In theses cases, the "factoring ac method" may be inconvenient due to large numbers of the product ac. However, The Diagonal Sum Method can transform a multiple steps solving process into a simplified one by doing some elimination operations.

Example 7. Solve: 12x^2 - 32x + 21 = 0.

1. Diagonal Sum method. Both roots are positive. Write the (c/a) setup:
Numerator: (1, 21),(3, 7).
Denominator: (1, 12),(2, 6),(3, 4).

First eliminate the pair (1, 21)/(1, 12) because it gives large diagonal sums (while b = 32). Then eliminate the pair (3, 4) since it gives odd-number diagonal sums (while b is even number). The remainder (c/a) is: (3, 7)/(2, 6) that leads to 2 probable root pairs: (3/2, 7/6) and (3/6, 7/2). The diagonal sum of first pair is: 18 + 14 = 32 = -b. The 2 real roots are 3/2 and 7/6.

2. The "ac method". Find 2 number: product is ac = 252; and sum is b = -32. Proceeding: [(-1, -252),(1, 252),(-2, -126)92, 126).........(-12, -21),(12, 21), (-14, -18), OK]. Next, replace the term (-32x) by the 2 terms (-14x) and (-18x) in the inequality:
12x^2 - 14x - 18x + 21 = 6x(2x - 3) - 7(2x - 3) = (2x - 3)(6x - 7) = 0.
(2x - 3) = 0 ; ---> x = 3/2 ; (6x - 7) = 0 ; ---> x = 7/6.

Example 8. Solve: 12x^2 + 5x - 72 = 0.

1. Diagonal Sum Method. Roots have opposite signs. The (c/a) setup;

Numerator: (-1, 72),(-2, 36),(-3, 24),(-4, 18),(-6, 12),(-8, 9).
Denominator: (1, 12),(2, 6),(3,4).

Before proceeding solving, look to eliminate the pairs that are not fitted. First, eliminate the pairs from the numerator: (-2, 36),(-4, 18),(-6, 12), and the pair (2, 6) from the denominator, because they will give even-number diagonal sums (while b is odd).
Next, eliminate the pairs: (-1, 72),(-3, 24)/(1, 12) because they give large diagonal sums (while b = 5). 
The remainder (c/a) is: (-8, 9)/(3, 4) that gives 2 probable root pairs: (-8/3, 9/4) and (-8/4, 9/3). The diagonal sum of first pair is: 27 - 32 = -5 = -b. The 2 real roots are -8/3 and 9/4.

2. The factoring "ac method". Find 2 numbers: product ac = -864; sum = 5. Since the product ac is a large number, proceeding is complicated. Proceed:[(-1, 864)(1, -864)...........(-18, 48)(18, -48)(-24, 36)(24, -36)(-32, 27)(32, -27) OK].
12x^2 - 27x + 32x - 72 = 3x(4x - 9) + 8(4x - 9) = (4x - 9)(3x + 8) = 0.
(4x - 9) = 0 ---> x = 9/4 ; (3x + 8) = 0 ---> x = -8/3.

Example 9. Solve: 24x^2 + 59x + 36 = 0.

1. Diagonal Sum method. Both roots are negative. Write the (c/a) setup:

Numerator: (-1, -36),(-2, -18),(-3, -12),(-4, -9),(-6, -6).
Denominator: (1, 24),(2, 12),(3, 8),(4, 6).

Before proceeding, eliminate the pairs: (-2, -18),(-6, -6)/(2, 12),(4, 6) because they give even-number diagonal sums (while b is odd). Also, eliminate the pairs: (-1, -36),(-3, -12)/(1, 24) because they give large diagonal sums (while b = 59).

The remainder (c/a) is : (-4, -9)/(3, 8). It gives 2 probable root pairs: (-4/3, -9/8) and (-4/8, -9/3). The diagonal sum of first pair is: -32 - 27 = -59 = -b. The 2 real roots are -4/3 and -9/8.

2. The "ac method". Find 2 numbers: product ac = 864, sum = b = 59. since the product ac is a large number, proceeding takes a lot of time. Proceed [(-1, -864),(1, 864),(-2, -432),(2, 432).......(-24, -36),(24, 36),(-27, -32),(27, 32) OK]. Next, replace term 59x by 2 terms 27x and 32x in the equation: 24x^2 + 27x + 32x + 36 = 8x(3x + 4) + 9(3x + 4) = (3x + 4)(8x + 9) = 0.
(3x + 4) ---> x = -4/3 ; (8x + 9) = 0 ---> x = -9/8.

Example 10. Solve 45x^2 + 74x + 24 = 0.

1. The Diagonal Sum method. Both roots are negative. Create the (c/a) setup:
Numerator: (-1, -24),(-2, -12),(-3, -8),(-4, -6).
Denominator: (1, 45),(3, 15)(5, 9).

First, eliminate the pairs (-1, -24),(-2, -12)/(1, 45), (3, 15) because they give large diagonal sums, as compared to b = 74. Then, eliminate the pair (-3, -8) because it gives an odd-number diagonal sum (while b = 74 is even). The remainder (c/a) is: (-4, -6)/(5, 9) that leads to 2 probable root pairs: (-4/6, -6/9) and (-4/9, -6/5). The diagonal sum of the second pair is: -54 - 20 = -74 = -b. The 2 real roots are -4/9 and -6/5.

2. The "ac method". Find 2 numbers, knowing ac = 1080 and b = 74. Proceeding:[(-1, -1080),(1, 1080),
(-2, -540),(2, 540)........(-20, -54),(20, 54),OK]. Replace in the equation 74x by 20x and 54x.
45x^2 + 20x + 54x + 24 = 5x(9x + 4) + 6(9x + 4) = (9x + 4)(5x + 6) = 0.
(9x + 4) = 0  ---> x = -4/9 ; (5x + 6) = 0 ---> x = -6/5.

CONCLUSION AND REMARKS.

1. Both methods deserves to be studied because they provide students with opportunities for improving math skills and logical thinking that are the ultimate goals of learning math.

2. When the constant a and c are large numbers and may contain themselves many factors, it is advised that students use the quadratic formula for solving. However, either doing it mentally or by calculator, remember to always proceed solving in 2 steps. 
First step, compute by calculator the Discriminant D = b^2 - 4ac. If the given quadratic equation can be factored, D must be a perfect square, and D's square root must be a whole number.
In second step, you should algebraically calculate the rest of the formula with the result (whole number) from step 1. Make sure that the answers be in the form of 2 fractions, and not in decimals.

3. If you master the Diagonal Sum Method, you may find it easier and more convenient to solve these cases than using the quadratic formula, especially when calculators are not available.