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

Apr 17, 2015

Say for instance you have...

${x}^{2} + b x$

This can be transformed into:

${\left(x + \frac{b}{2}\right)}^{2} - {\left(\frac{b}{2}\right)}^{2}$

Let's find out if the expression above translates back into ${x}^{2} + b x$...

${\left(x + \frac{b}{2}\right)}^{2} - {\left(\frac{b}{2}\right)}^{2}$

$= \left(\left\{x + \frac{b}{2}\right\} + \frac{b}{2}\right) \left(\left\{x + \frac{b}{2}\right\} - \frac{b}{2}\right)$

$= \left(x + 2 \cdot \frac{b}{2}\right) x$

$= x \left(x + b\right)$

$= {x}^{2} + b x$

Now, it is important to note that ${x}^{2} - b x$ (notice the minus sign) can be transformed into:

${\left(x - \frac{b}{2}\right)}^{2} - {\left(\frac{b}{2}\right)}^{2}$

What you are doing here is completing the square . You can solve many quadratic problems by completing the square.

Here is one primary example of this method at work:

$a {x}^{2} + b x + c = 0$

$a {x}^{2} + b x = - c$

$\frac{1}{a} \cdot \left(a {x}^{2} + b x\right) = \frac{1}{a} \cdot - c$

${x}^{2} + \frac{b}{a} \cdot x = - \frac{c}{a}$

${\left(x + \frac{b}{2 a}\right)}^{2} - {\left(\frac{b}{2 a}\right)}^{2} = - \frac{c}{a}$

${\left(x + \frac{b}{2 a}\right)}^{2} - {b}^{2} / \left(4 {a}^{2}\right) = - \frac{c}{a}$

${\left(x + \frac{b}{2 a}\right)}^{2} = {b}^{2} / \left(4 {a}^{2}\right) - \frac{c}{a}$

${\left(x + \frac{b}{2 a}\right)}^{2} = {b}^{2} / \left(4 {a}^{2}\right) - \frac{4 a c}{4 {a}^{2}}$

${\left(x + \frac{b}{2 a}\right)}^{2} = \frac{{b}^{2} - 4 a c}{4 {a}^{2}}$

$x + \frac{b}{2 a} = \pm \frac{\sqrt{{b}^{2} - 4 a c}}{\sqrt{4 {a}^{2}}}$

$x + \frac{b}{2 a} = \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$

$x = - \frac{b}{2 a} \pm \frac{\sqrt{{b}^{2} - 4 a c}}{2 a}$

$\therefore x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

The famous quadratic formula can be derived by completing the square .

Apr 18, 2015

The new Transforming Method to solve quadratic equations.
CASE 1. Solving type ${x}^{2} + b x + c = 0$. Solving means finding 2 numbers knowing their sum ($- b$) and their product ($c$). The new method composes factor pairs of ($c$), and in the same time, applies the Rule of Signs. Then, it finds the pair whose sum equals to ($b$) or ($- b$).
Example 1. Solve ${x}^{2} - 11 x - 102 = 0$.
Solution. Compose factor pairs of $c = - 102$. Roots have different signs. Proceed: $\left(- 1 , 102\right) \left(- 2 , 51\right) \left(- 3 , 34\right) \left(- 6 , 17\right) .$ The last sum $\left(- 6 + 17 = 11 = - b\right) .$ Then the 2 real roots are: $- 6$ and $17$. No factoring by grouping.
CASE 2 . Solving standard type: $a {x}^{2} + b x + c = 0$ (1).
The new method transforms this equation (1) to: ${x}^{2} + b x + a \cdot c = 0$ (2).
Solve the equation (2) like we did in CASE 1 to get the 2 real roots ${y}_{1}$ and ${y}_{2}$. Next, divide ${y}_{1}$ and ${y}_{2}$ by the coefficient a to get the 2 real roots ${x}_{1}$ and ${x}_{2}$ of original equation (1).
Example 2. Solve $15 {x}^{2} - 53 x + 16 = 0$. (1) $\left[a \cdot c = 15 \left(16\right) = 240\right] .$
Transformed equation: ${x}^{2} - 53 + 240 = 0$ (2). Solve equation (2). Both roots are positive (Rule of Signs). Compose factor pairs of $a \cdot c = 240$. Proceed: $\left(1 , 240\right) \left(2 , 120\right) \left(3 , 80\right) \left(4 , 60\right) \left(5 , 48\right)$. This last sum is $\left(5 + 48 = 53 = - b\right)$. Then, the 2 real roots are: ${y}_{1} = 5$ and
${y}_{2} = 48$. Back to original equation (1), the 2 real roots are: x_1 = y_1/a = 5/15 = 1/3; and ${x}_{2} = {y}_{2} / a = \frac{48}{15} = \frac{16}{5.}$ No factoring and solving binomials.

The advantages of the new Transforming Method are: simple, fast, systematic, no guessing, no factoring by grouping and no solving binomials.