Given that the equation $$kx^2-2x+3-2k=0$$ has equal roots, find the possible values of the constant $k$.
So far I have only attempted to factor $-2$ which I'm not sure is even right. I understand I need to use the discriminant later on in the Question. Completely lost, any guidance would be helpful. If you post the solution please show it step by step.
Thanks!
Best Answer
The discriminant of a quadratic $ax^2+bx+c$ is given by $\Delta=b^2-4ac.$ Moreover the quadratic has equal roots when $\Delta=0$.
In this case the quadratic is $kx^2-2x+3-2k$, so $b=-2$, $a=k$ and $c=3-2k$.
Can you end it?
Also by the quadratic equation we have $$x=\frac{2\pm\sqrt{\Delta}}{2k}=\frac{2\pm\sqrt{4-4k(3-2k)}}{2k}$$ $$=\frac{2\pm\sqrt{8k^2-12k+4}}{2k}$$ $$=\frac{2\pm2\sqrt{2k^2-3k+1}}{2k}=\frac{1\pm\sqrt{2k^2-3k+1}}{k}$$
Since you want equal roots (a repeated root), we solve $2k^2-3k+1=0$ for $k$.