Question

What is the new Transforming Method to solve quadratic equations?

Answer 1

The new Transforming Method to solve quadratic equations

Explanation:

1. Solving equation type `x^2 + bx + c = 0` with a = 1.
   Solving is finding 2 numbers knowing sum (-b) and product (c). Compose factor pairs of c by observing the Rule of Signs. Then, find the pair whose sum is (-b).
   Example 1: Solve x^2 - 31x - 102 = 0. Roots have different signs.
   Compose factor pairs of c = -102--> (-2, 51)(-3, 34). This sum is 31 = -b. Then the 2 real roots are: (-3 and 34). No factoring, no solving binomials.
   Example 2. Solve x^2 - 28x + 96 = 0. Both roots are positive.
   Factor pairs of c = 96 -> (2, 48)(3, 32)(4, 24). This sum is 28 = -b. Then the 2 real roots are: 4 and 24.
   Example 3. Solve x^2 + 39x + 108 = 0. Both roots are negative.
   Factor pairs of c = 108 -> (-2, -54)(-3, -36). This sum is -39 = -b. Then the 2 real roots are: -3 and - 36.
   We will see Case 2, solving standard type ax^2 + bx + c = 0 in the next explanation.

Answer 2

1. Solving standard type `y = ax^2 + bx + c = 0` (1)
   Bring this case to Case 1 by transforming the equation (1) to->
   `y' = x^2 + bx + a.c = 0` (2)
   Find the 2 real roots y1 and y2 of equation (2). The 2 real roots of original equation (1) are: `x1 = (y1)/a and x2 = (y2)/a`.
   Example 1. Solve `y = 8x^2 - 22x - 13 = 0` (1).
   Transformed equation `y' = x^2 - 22x - 104 = 0` (2). Roots have different signs.
   Factor pairs of ac = -104 --> (-2, 53)(-4, 26). This sum is 22 = -b. The
   2 real roots of (2) 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 2. Solve `y = 24x^2 + 59x + 36 = 0` (1)
   Transformed equation: `y' = x^2 + 59x + 864 = 0` (2). Both roots are negative.
   Factor pairs of ac = 864 --> ...(-18, -48)(-24, -36)(-27, -32). This sum is
   -59 = -b. Then y1 = -27 and y2 = -32. Then, `x1 = -27/24 = -9/8` and `x2 = -32/24 = -4/3`.