How to solve quadratic equations by the new Diagonal Sum Method?

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justaguide | College Teacher | (Level 2) Distinguished Educator

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For a quadratic equation ax^2 + bx + c = 0, first find the signs of the roots using the following rules:

1. If a and c have opposite signs, the roots have opposite signs.

2. If a and c have the same sign, the roots have the same sign.

Also, when this is the case, if a and b have opposite signs, the roots are positive, else they are negative.

Next, using the Diagonal Sum Method of solving quadratic equations the denominator of the roots is taken as the factor set of a and the numerator of the roots is taken as the factor set of b.

Using this a probable list of roots is created. To choose the correct roots from this, the sum of the product of the numerator and the denominator of the roots should be equal to -b.

For example take the quadratic equation: x^2 + 4x + 4 = 0. As a and c have the same sign, the roots have th same sign and as a and b have the same sign they are negative.

Now the factors of a are 1 and the factors of c are 1, 2 and 4

Using these we can create the probable root sets: (-4/1, -1/1) and (-2/1 and -2/1)

Taking the diagonal sum: -5, -4

As -4 = -b, the roots are (-2, -2)

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ngkimthu | Student, Undergraduate | (Level 1) eNoter

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Solving complicated quadratic equations by the diagonal sum method.

Examples of quadratic equations that are considered complicated in solving: 24x^2 + 59x + 36 = 0; 12x^2 +5x - 72 = 0; 45x^2 - 74x - 55 = 0...

Solving these equations by the quadratic formula may have some computation difficulties, especially when you can't use calculators, during some tests/exams, for example. Solving them by the factoring method also takes a lot of time due to the high number of permutations. Below is a suggested approach to solve these equations using the diagonal sum method.

Solving process. First, create an all-options setup (c/a) that includes all probable root-pairs. In this setup, the numerator contains all factor-pairs of (c). The denominator contains all factor-pairs of (a).

Example. The all-options setup of the equation 12x^2 - 272x + 45 = 0 contains:

Numerator: All factor-pairs of c = 45: (1, 45)(3, 15)(5, 9)   Denominator: All factor-pairs of a = 12:   (1, 12)(2, 6)(3, 4)

Next, proceed the elimination of the options that do not fit. The remainder options usually requires fewer than 3 trials in order to get the 2 real roots.

Practical approaches to get rid of non-fitted options:

1. Eliminate options that give extreme DS (larger or smaller) as compared with (b).

2. If b is odd number, eliminate options that give even-number DS, meaning the ones that link to such pairs like (2, 6), (4, 8), (6, 10)...

3. If b is even-number, eliminate options that give odd-number DS, meaning the ones that link to such pairs like (1, 4), (3, 8), (5, 6)...

Examples of solving.

Example 1. Solve 12x^2 - 272x + 45 = 0. Both roots are positive. Create the all-option setup:

Numeratot:  (1, 45)(3, 15)(5, 9)

Denominator: (1, 12)(2, 6)(3, 4)

First, eliminate the options linked to pairs: (1, 12), (3, 4) since they give odd-number DS (while b is even). Next, look for the option that gives a large DS (272). That would be the option (1, 45)/((2, 6) that gives 2 probable rootpairs: (1/2, 45/6) and (1/6, 45/2). The second DS is 2 + 270 = 272 = -b. The 2 real roots are 1/6 and 45/2.

Example 2. Solve: 12x^2 + 5x - 72 = 0. Roots have opposite signs. Create all option setup:

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

(1, 12)(2, 6)(3, 4)

First, eliminate the options linked to pairs (-2, 36)(-4, 18)(-6, 12)/(2, 6) since they give even-number DS (while b is odd). Then eliminate options linked to pairs ((-1, 72)(-3, 24)/(1, 12) since they give large DS (while b  = 5). The remainder option (-8, 9)/(3, 4) gives 2 probable root pairs:(-8/3, 9/4) and (-8/4, 9/3). The first DS is 27 - 32 = -5 = -b. The 2 real roots are -8/3 and 9/4.

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ngkimthu | Student, Undergraduate | (Level 1) eNoter

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The Diagonal Sum Method for solving quadratic equations.

Concept: Direct finding 2 real roots in the form of 2 fractions knowing their sum (-b/a) and product (c/a).

Recall the rule of signs for real roots. Examples:

1. 4x^2 + 24x - 13 = 0 has 2 roots with opposite signs

2. 15x^2 + 22x + 8 = 0 has 2 real roots both negative.

3. 3x^2 - 13x + 10 = 0 has 2 real roots both positive.

The Diagonal Sum of a root pair.

Given a pair of real roots: (2/3 , 4/5). its DS is 10 + 12 = 22.

Rule for the Diagonal Sum. The DS of a true root pair must equal to (-b). If it equals to (b), then it is the negative of the solution.

Example: For the DS (22) and the root pair (2/3, 4/5) above, the equation they justify is 15x^2 - 22x + 8 = 0. Example: The equation 15x^2 - 7x + 4 = 0 has one real root pair (-1/3 , 4/5) and its DS is -5 + 12 = 7 = -b.

SOLVING PROCESS. The DSM directly selects the probable root pairs, based on the product (c/a). Then, it uses mental math to calculate their diagonal sums to find the one that is equal to -b.

SOLVING PRACTICES.

A. When a =1 - Solving x^2 + bx + c = 0. The DS becomes the sum of the 2 real roots. solving doesn't need factoring.

Example: Solve x^2 - 39 x + 108 = 0. Both real roots are positive. Write factor pairs of c = 108. They are (1, 108) (2, 54) (3, 36)... Stop! This sum is 39 = -b. The 2 real roots are 3 and 36.

B. When a and c are prime numbers. The number of probable root pairs is limited to one, except when 1 (or -1) is a real root.

Example. Solve: 7x^2 - 118x - 17 = 0. Roots have opposite signs. There is unique root pair: (-1/7 , 17/1). Its DS is -1 + 119 = 118 = -b. The 2 real roots are -1/7 and 17.

C. When a and c are small numbers and may have themselves one factor. The DSM write down all probable root pairs, the use mental math to find the DS that fits.

Example. Solve: 7x^2 - 57x + 8 = 0. Both roots are positive. Probable root pairs: (-1/7, 8/1) (2/1, 4/7) (2/7, 4/1). The DS of the first pair is 57 = -b. The 2 real roots are 1/7 and 8.

Example. Solve: 6x^2 - 19x - 11 = 0. Roots have opposite signs. Write all probable root pairs: (-1/6, 11/1) (-1/2, 11/3) (-1/3, 11/2). The DS of the second set is 19 - -b. The 2 real roots are -1/2 and 11/3.

D. When a and c are large numbers and may have themselves many factors. Solving these quadratic equations is considered complicated since it involves many permutations. To know how to solve them, see the article titled:"How to solve complicated quadratic equations" on the WikiHow or Bukisa websites.

 

 

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ngkimthu | Student, Undergraduate | (Level 1) eNoter

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Solving quadratic equations by the Diagonal Sum Method.

A When a = 1, the diagonal sum reduces to the sum of the 2 real roots. Solving is simple and doesn't need factoring.

Example 1. Solve x^2 - 60x + 116 = 0. Solution. The rule of signs shows that the 2 roots are both positive. Factor-sets of c = 116: (1, 116), (2, 58)...stop. This sum is 2 + 58 = 60 = -b. The 2 real roots are 2 and 58.

Example 2. Solve: x^2 + 37x - 120 = 0. Solution. Roots have opposite signs. Factor-sets of c = -120: (-1, 120), (-2, 60),(-3, 48)...stop. This sum is -3 + 40 = 37 = b. Rule of signs indicates the answer is the negative of this set. The 2 real roots are 3 and -40.

Example 3. Solve x^2 + 51x + 144 = 0. Solution. Both roots are negative. Factor sets of c = 144: (-1,-144),(-2,-72), (-3,-48)..... Stop. This sum is -3 - 48 = -51 = -b. The real roots are -3 and -48.

B. When a and c are prime numbers. The number of probable root-sets is limited to 1.

Example 4. Solve 5x^2 - 64x - 13 = 0. Solution. Roots have opposite signs. There is unique root-set: (-1/13 , 5/1). Diagonal sum: 64 = -b. The real roots are -1/13 and 5.

Example 5. Solve: 7x^2 - 78x + 11 = 0. Solution. Both roots are positive, Unique root-set: (1/7 , 11/1). Diagonal sum: 78 = -b. The real roots are 1/7 and 11.

C. When a and c are small numbers and contain themselves one factor. Number of probable root-sets are fewer than 4.

Example 6. Solve: 6x^2 - 113x - 19 = 0. Solution. Roots have opposite signs. a = 6 has 2 factors. There are 3 probable root-sets:(-1,/2 , 19/3) , (-1/3 , 19/2), (-1/6 , 19/1). Diagonal sum of third set: 113 = -b. The real roots are -1/6 and 19.

Example 7. Solve 13x^2 + 33x + 14 = 0. Solution. Both roots are negative. Probable root-sets: (-1/13, -14/1),(-2/1, -7/13),(-2/13, -7/1). Diagonal sum of second set: -7 - 26 = -33 = -b. The real roots are -2 and -7/13.

D. Solving complicated quadratic equations. They are considered complicated because both a and c are large numbers and contain themselves many factors. There are many permutations involved. Examples: 45x^2 - 172x + 36 = 0; 12x^2 - 272x + 45 = 0; 24x^2 +59x + 36 = 0. However, through elimination process, we can limit the number of probable root-sets to fewer than 3 before proceed solving. Please read articles on Bukisa.com and wikihow.com.

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