What is the new Transforming Method to solve quadratic equations?

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What is the new Transforming Method to solve quadratic equations?

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Answer 1 · 2015-04-17T23:44:35

Say for instance you have...

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

This can be transformed into:

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

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

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

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

$= (x + 2 \cdot \frac{b}{2}) x$ $=(x+2*b/2)x$

$= x (x + b)$ $=x(x+b)$

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

The answer is YES.

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

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

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:

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

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

$\frac{1}{a} \cdot (a x^{2} + b x) = \frac{1}{a} \cdot - c$ $1/a*(ax^2+bx)=1/a*-c$

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

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

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

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

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

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

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

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

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

$:.x=(-b+-sqrt(b^2-4ac))/(2a)$

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

Answer 2 · 2015-04-18T23:21:03

The new Transforming Method to solve quadratic equations.
CASE 1. Solving type $x^2 + bx + c = 0$ . 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 $x^2 - 11x - 102 = 0$ .
Solution. Compose factor pairs of $c = -102$ . Roots have different signs. Proceed: $(-1, 102)(-2, 51)(-3, 34)(-6, 17).$ The last sum $(-6 + 17 = 11 = -b).$ Then the 2 real roots are: $-6$ and $17$ . No factoring by grouping.
CASE 2 . Solving standard type: $ax^2 + bx + c = 0$ (1).
The new method transforms this equation (1) to: $x^2 + bx + a*c = 0$ (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 $15x^2 - 53x + 16 = 0$ . (1) $[a*c = 15(16) = 240].$
Transformed equation: $x^2 - 53 + 240 = 0$ (2). Solve equation (2). Both roots are positive (Rule of Signs). Compose factor pairs of $a*c = 240$ . Proceed: $(1, 240)(2, 120)(3, 80)(4, 60)(5, 48)$ . This last sum is $(5 + 48 = 53 = -b)$ . 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 $x_2 = y_2/a = 48/15 = 16/5.$ 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.

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What is the new Transforming Method to solve quadratic equations?

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