Find k for which the equation has equal roots.

Question

I find to find the value of k in terms of $\alpha$ and $\beta$ . I rearranged the equation $x^2 +kx-1=2k+x$ into $x^2+(k-1)x-(1+2k)=0.$

I then found $\alpha$$\beta$ to equal$-1-2k$. and $\alpha + \beta$ to equal $-k+1.$

I was then asked to find $k$ for which the equation has equal roots. I am unsure of what to do. Does this just mean when $\alpha$ = $\beta$ ? And how do I incorporate my previously found information into it?

Thank you, all help is appreciated.

Answer 1

If you don't want to incorporate your prior knowledge to the problem, the quadratic equation has equal roots when the discriminant is equal to zero. $$(k-1)^2 + 4(1)(1+2k) = k^2 + 6k + 5 = (k+1)(k+5) = 0 \iff k \in \{ -5, -1 \} $$

Answer 2

Equal roots must satisfy the condition: $b^2 -4ac = 0$

You need to solve:

$$ (k-1)^2 -4(-1-2k) = 0 $$