What is the new Transforming Method to solve quadratic equations?

1 Answer
Aug 20, 2015

Solving quadratic equations by the new Transforming Method
(Google, Yahoo, Bing Search)

Explanation:

Case 1 . Equation type: x^2 + bx + c = 0x2+bx+c=0, with a = 1.
Find 2 numbers knowing sum (-b) and product (c). Compose factor pairs of c, and in the same time apply the Rule of Signs. Find the factor pair that equals to b, or (-b).
Example 1. Solve x^2 - 11x - 102 = 0x2โˆ’11xโˆ’102=0.
Roots have opposite signs. Factor pairs of (-102)--> ...(-2, 51)(-3, 34)(-6, 17). This sum is 11 = -b. Then, the 2 real roots are: -6 and 17.
Example 2. Solve x^2 - 27x + 126x2โˆ’27x+126.
Both roots are positive, Factor pairs of 126 --> ...(2, 63)(3, 42)(6, 21). This sum is 27 = -b. Then, the 2 real roots are: 6 and 21
Case 2 . Equation standard type: y = ax^2 + bx + c = 0y=ax2+bx+c=0 (1)
Transformed equation: y' = x^2 + bx + ac = 0 (2).
Solve the equation (2) exactly like in Case 1. Find its 2 real roots y1 and y2. Next, find the 2 real roots of the original equation (1) by these relations: x1 = (y1)/a and x2 = (y2)/a.
Example 3. Solve 8x^2 - 22x - 13 = 0 (1).
Transformed equation: y' = x^2 - 22x - 104. Roots have opposite signs. Factor pairs of (ac = - 104) --> (-2, 52)(-4, 26). This sum is 22 = -b. The 2 real roots are: y1 = -4 and y2 = 26.
Back to original equation (1), the 2 real roots are: x1 = (y1)/a = -4/8 = -1/2, and x2 = (y2)/a = 26/8 = 13/4.
Example 4. Solve y = 12x^2 + 29x + 15 = 0 (1)
Transformed equation: y' = x^2 + 29x + 180 = 0 (2). Both roots are negative. Factor pairs of ac = 180 --> (-6, -30)(-9, -20). This sum is -29 = -b. Then y1 = -9 and y2 = -20. Back to (1), the 2 real roots are: x1 = (y1)/a = -9/12 = -3/4, and x2 = (y2)/a = -20/12 = -5/3
NOTE. This method is fast, systematic, no guessing, no lengthy factoring by grouping.