Reverse Engineering a Quadratic

Nothing new, just something I found kind of fun.

Let's solve this quadratic through factorisation: \(x^2-6x+8=0\).

Appropriate factors are -4 and -2: \(x^2-4x-2x+8=0\)

\(x(x-4)-2(x-4)=0\)

\((x-2)(x-4)=0\)

\(x-2=0, x-4=0\)

\(x=2\) or \(x=4\)

Now, let's start out with the roots of this quadratic and form an equation.

\(x=2\) or \(x=4\)

\((x-2)(x-4)=0\)

\(x^2-2x-4x+8=0\)

\(x^2-6x+8=0\). We're back at the original equation.

Let's try one more: \(x=-4\) and \(x=1\).

\((x+4)(x-1)=y\)

\(x^2+4x-x-4=y\)

\(x^2+3x-4=y\)

\(y=x^2+3x-4\)

This article was written on 18/09/2023. If you have any thoughts, feel free to send me an email with them. Have a nice day!