In $2x^2-4x-k=0$, $k_1$ represents the value of $k$ if the two roots are equal, $k_2$ if one root is 10 and $k_3$ if one exceeds the other by 10. Find $k_1+k_2+k_3$.
A hint is to use sum of roots theorem and product of roots theorem. I need to find the possible roots so that it meets the rules so I can add them.
The product of the roots is $-\frac k2$ and their sum is $\frac{-4}{-2}=2$.
For $k_1$, both roots must be 1, so $-\frac{k_1}2=1\cdot1\implies k_1=-2$.
For $k_2$, the other root is $-8$, so $-\frac{k_2}2=10\cdot-8\implies k_2=160$.
For $k_3$, one root is $-4$ and the other is 6, so $-\frac{k_3}2=-4\cdot6\implies k_3=48$.
Their sum is $k_1+k_2+k_3=206$.