# How to Solve Complex Cases of Quadratic Equations

|right|251px]]The standard form of a quadratic equation in one variable is ax² + bx + c = 0. Depending on the values of the constants a and c, solving this equation may be simple or may be complicated. To know about the new Diagonal Sum Method, please read the article:"How to solve quadratic equations by the Diagonal Sum Method" on this website. This article discusses various cases in solving quadratic equations.

### Method 1 VARIOUS CASES IN SOLVING QUADRATIC EQUATIONS

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A. When a = 1 - Solving quadratic equations types x² + bx + c = 0.
• Solving this type of quadratic equations results in solving a popular puzzle: finding two numbers knowing their sum and their product. Solving becomes simple and doesn't need factoring.
• Example 1. Solve: x² - 26x - 72 = 0.
• Solution. Both real roots have opposite signs. Write down the factor-pairs of c = -72. They are:
• (-1 , 72)(-2 , 36)(-3 , 24)...Stop!The sum of the 2 real roots in this set is 21 = -b. The 2 real roots are -3 and 24.
• Example 2. Solve: -x² - 26x + 56 = 0.
• Solution. Roots have opposite signs. Write down factor-pairs of c = 56:
• (-1, 56) (-2, 28)...Stop!. This sum is 26 = b. According to the Diagonal Sum Rule, when a is negative, the answers are -2 and 28.
• Example 3. Solve x² + 27x + 50 = 0.
• Solution. Both real roots are negative. Write factor-sets of c = 50:
• (-1, -50) (-2, -25)..Stop! This sum is -27 = -b. The 2 real roots are -2 and -25.
• Example 4. Solve: x² - 39x + 108 = 0.
• Solution. Both real roots are positive. Write the factor-sets of c = 108:
• (1, 108) (2, 54) (3, 36)...Stop! This sum is 39 = -b. The 2 real roots are 3 and 36.

### Method 2 B. When a and c are prime/small numbers.

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In these cases, the new method directly selects the probable root pairs basing on the (c/a) setup and by, in the same time, applying the rule of signs.
• Example 5. Solve: 7x² + 90x - 13 = 0.
• Solution. Roots have opposite signs. Both a and c are prime numbers. Since 1 (or -1) is not a real root, there is a unique root pair:
• (-1/7, 13/1). Since its diagonal sum is: -1 + 91 = 90 = b, the answers are opposite in sign. The 2 real roots are 1/7 and -13.
• Example 6. Solve: 7x² - 57x + 8 = 0.
• Solution. Both roots are positive. The c/a setup: (1, 8)(2, 4)/(1, 7).
• Probable root pairs: (1/7 , 8/1) (2/1 , 4/7) (2/7 , 4/1)
• Diagonal Sum of first set:(1 + 56) = 57 = -b.
• The 2 real roots are 1/7 and 8.
• NOTE. You can simplify the c/a setup before proceeding. Eliminate the pair (2, 4) from the numerator because it gives even-number diagonal sums (while b is odd). The remainder c/a: (1, 8)/(1, 7) leads to unique probable root pair: (1/7, 8/1) that gives the 2 real roots 1/7 and 8.
• Example 7. Solve: 6x² - 19x - 11 = 0.
• Solution. Roots have opposite signs. Write the c/a setup:
• (-1, 11)/(1, 6)(2, 3)
• Eliminate the pair (1, 6) because it gives larger diagonal sum (while b = -19).
• The remainder c/a: (-1, 11)/(2, 3) leads to 2 probable root pairs: (-1/2, 11/3) and (-1/3, 11/2). The first diagonal sum is: -3 + 22 = 19 = -b. The 2 real roots are -1/2 and 11/3.

### Method 3 C. COMPLICATED CASES OF QUADRATIC EQUATIONS

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When the constants a and c are large numbers and may contain themselves many factors, solving is considered complicated because there are many permutations involved. The Diagonal Sum Method proceeds basing on the (c/a) setup. The numerator of the setup contains all factor pairs of c. The denominator contains all factor pairs of a.
• Next, this new method transforms a multiple steps solving process into a simplified one by doing a few operations of elimination.
• Example 1. Solve: 45x² - 74x - 55 = 0.
• Solution. Roots have opposite signs. Write down the (c/a) setup.
• Numerator. All factors-sets of c: (-1, 55) (-5, 11).
• Denominator. All factors-sets of a: (1, 45) (3, 15) (5, 9).
• Smart students can use mental math to calculate all diagonal sums and find the one that fits.
• The best way is to proceed elimination of the factor pairs that do not fit.
• Eliminate the pairs: (-1, 55)/(1, 45)(3, 15) since they give larger diagonal sums (while b = -74).
• The unique remainder (c/a) is: (-5, 11)/(5, 9) that gives as 2 real roots: -5/9 and 11/5.
• Example 2. Solve: 12x² - 272x + 45 = 0.
• Solution. Both roots are positive. Write down the (c/a) setup.
• Numerator: (1, 45) (3, 15) (5, 9)
• Denominator: (1, 12) (2, 6) (3, 4)
• First eliminate the pairs with (1, 12),(3, 4) since they will give odd-number diagonal sums (while b is an even number).
• Next, look for a setup (c/a) that gives a large diagonal sum. It should be the setup:(1, 45)/(2, 6) that leads to 2 probable root pairs:
• (1/2 , 45/6) and (1/6, 45/2). The second diagonal sum is: 2 + 270 = 272 = -b. The 2 real roots are 1/6 and 45/2.
• Example 3. Solve: 40x² - 483x + 36 = 0.
• Solution. Both roots are positive. Write down the c/a setup.
• Numerator: (1, 36) (2, 18) (3, 12) (6, 6).
• Denominator: (1, 40) (2, 20) (4, 10)
• First eliminate the pairs (2, 18)(6, 6)/(2, 20)(4, 10) because they give even-number diagonal sums (while b is odd).
• The remainder c/a leads to 3 probable root pairs: (1/40, 36/1), (3/1, 12/46), (3/40, 12/1). The third diagonal sum is : 3 + 480 = 483 = -b. The 2 real roots are 3/40 and 12.
• Example 4. Solve: 12x² + 5x - 72 = 0.
• Solution. Roots have opposite signs. Write the (c/a) setup.
• Numerator: (-1, 72)(-2, 36)(-3, 24)(-4, 18)(-6, 12) (-8, 9)
• Denominator: (1, 12) (2, 6) (3, 4).
• First, eliminate the pairs (-2, 36),(-4, 18),(-6, 12)/(2, 6) since they give even-number diagonal-sums (while b is odd).
• Next, eliminate the pairs (-1, 72),(-3, 24)/(1, 12) since they give large diagonal sums ( while b is small).
• The remainder (c/a) is (-8, 9)/(3, 4) that gives as 2 real roots: -8/3 and 9/4.
• Example 5. Solve: 24x² - 59x + 36 = 0.
• Solution. Both roots are positive. Write the (c/a) setup.
• Numerator: (1, 36) (2, 18) (4, 9) (6, 6)
• Denominator: (1, 24) (2, 12) (4, 6) (8, 3)
• First, eliminate the pairs (2, 18),(6, 6)/(2, 12),(4, 6) that give even-number diagonal sums (while b is odd).
• Next, eliminate the pairs (1, 36)/(1, 24) that gives large diagonal sums (while b = -59).
• The remainder c/a is (4, 9)/(8, 3) that gives as 2 real roots: 4/3 and 9/8.

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Categories: Algebra