The quadratic equation \(x^2-4x-1=2k(x-5)\) has two equal roots. Calculate the possible values of \(k\).
We can rewrite \(x^2-4x-1=2k(x-5)\) as \[x^2-(2k+4)x+10k-1=0.\] This has two equal roots exactly when the discriminant of the equation is zero, that is, when \[(2k+4)^2-4(10k-1)=0.\] Multiplying out the brackets, we find that \[4k^2+16k+16-40k+4=0\] which simplifies to \[4k^2-24k+20=0.\] Dividing through by \(4\), we find \[k^2-6k+5=0\] which factorises to \[(k-5)(k-1)=0.\] Therefore the equation has two equal roots exactly when \(k=1\) or \(k=5\).
To visualise this, you could create a GeoGebra file to show this curve and line, with a slider for \(k\). Use this file to confirm our results above.
