When does $x^2-4x-1=2k(x-5)$ have equal roots?

When does $x^2-4x-1=2k(x-5)$ have equal roots?

Question

The quadratic equation $x^2-4x-1=2k(x-5)$ has two equal roots. Calculate the possible values of $k$.

Suggestion

Here is the curve $y=x^2-4x-1-2k(x-5)$.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/c6RP6YDJ/width/500/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

Solution

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.