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 \(x\).