# What is the new Transforming Method to solve quadratic equations?

Aug 20, 2015

Solving quadratic equations by the new Transforming Method

#### Explanation:

Case 1 . Equation type: ${x}^{2} + b x + c = 0$, 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 ${x}^{2} - 11 x - 102 = 0$.
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 ${x}^{2} - 27 x + 126$.
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: $y = a {x}^{2} + b x + c = 0$ (1)
Transformed equation: $y ' = {x}^{2} + b x + a c = 0$ (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: $x 1 = \frac{y 1}{a}$ and $x 2 = \frac{y 2}{a} .$
Example 3. Solve $8 {x}^{2} - 22 x - 13 = 0$ (1).
Transformed equation: $y ' = {x}^{2} - 22 x - 104$. 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: $x 1 = \frac{y 1}{a} = - \frac{4}{8} = - \frac{1}{2}$, and $x 2 = \frac{y 2}{a} = \frac{26}{8} = \frac{13}{4}$.
Example 4. Solve $y = 12 {x}^{2} + 29 x + 15 = 0$(1)
Transformed equation: $y ' = {x}^{2} + 29 x + 180 = 0$ (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: $x 1 = \frac{y 1}{a} = - \frac{9}{12} = - \frac{3}{4}$, and $x 2 = \frac{y 2}{a} = - \frac{20}{12} = - \frac{5}{3}$
NOTE. This method is fast, systematic, no guessing, no lengthy factoring by grouping.