I have not used this method enough to say for certain if this has any speed benefits. I did find it interesting nonetheless.
We're familiar with how we solve a quadratic through completing the square the standard way. Take \(x^2+8x-9=0\):
\(x^2+8x+(4)^2-9-(4)^2=0\)
\((x+4)^2-9-16=0\)
\((x+4)^2-25=0\)
\((x+4)^2=25\)
\(\sqrt{(x+4)^2}=\sqrt{25}\)
\(x+4=\pm5\)
\(x=\pm5-4\)
\(x=1\) or \(x=-9\)
\(x^2+8x-9=(ax+b)^2-r\)
\(x^2+8x-9=a^2x^2+2abx+b^2-r\)
Equation 1: \(a^2=1\) or simply \(a=1\)
Equation 2: \(2ab=8\)
Equation 3: \(b^2-r=-9\)
Plugging 1 into 2: \(2(1)b=8\)
\(2b=8\)
\(b-4\)
Plugging 2 into 3: \(4^2-r=-9\)
\(16-r=-9\)
\(-r=-25\) or simply \(r=25\)
Now we know \(a=1\), \(b=4\) and \(r=25\). We'll put this into \((ax+b)^2-r\):
\((x+4)^2-25\). We solve as normal from here.
Essentially, we create an equation where our original quadratic equals our desired final form. Then, we bring the final form into quadratic form, and compare both sides of the equation.
This article was written on 07/09/2023 and modified on 14/09/2023. If you have any thoughts, feel free to send me an email with them. Have a nice day!