Method: Completing the square

$$ax^2+bx+c=0$$

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

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

Method: Introducing z

$$ax^2+bx+c=0$$

$$x^2+\frac{b}{a}x+\frac{c}{a}=0$$ Let the two roots be $R$ and $S$. So,

$$(x-R)(x-S)=x^2-(R+S)x+RS=x^2+\frac{b}{a}x+\frac{c}{a}$$ Equating the coefficients of $x$,

$R+S=\frac{-b}{a}$

So, the mean of $R$ and $S$ is $\frac{-b}{2a}$

Now, two numbers have equal distance from their mean with different signs. So $R$ and $S$ have equal absolute distance from $\frac{-b}{2a}$. Let one distance be $z$. So the other distance is $-z$. So, $R$ and $S$ can be written as:

$$\frac{-b}{2a}\pm z$$ Now, we need to get the value of $z$.

Equating the products of two roots:

$$RS=(\frac{-b}{2a}+ z)(\frac{-b}{2a}- z)=\frac{c}{a}$$ $$\frac{b^2}{4a^2}-z^2=\frac{c}{a}$$ $$z^2=\frac{b^2}{4a^2}-\frac{c}{a}=\frac{b^2-4ac}{4a^2}$$ $$z=\pm\frac{\sqrt{b^2-4ac}}{2a}$$ Plugging the value of $z$, the roots are:

$$\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}$$

Comments

• More calculations need to be done in the new method.

• In the old method, the only technique to remember is the completion of square. For this, of course we need to remember the $(a+b)^2$ formula. Knowing the formula, completing the square part is more of a technique than to memorize. In the new method, along with coefficient equating method, we need to know that two numbers are equidistance (with opposite direction) from their mean.

• The classical method is straigtforward. Starting with the equation, it directly attempts to solve for $x$, which is what we want. The new method is not that straightforward. We are introducing $z$, and later solving for $z$ in order to get $x$.