I'm stuck on this question, and I have the answer but I don't know how to get to it. This is what I've done so far.
For the quadratic equation $ax^2+bx+c=0$ to have two real roots, the discriminant must be greater than $0$:
$$
b^2 – 4ac > 0.
$$
After substituting $a$, $b$ and $c$ in, I got
$$
2^2 – 4k(1-k) > 0.
$$
And then after expanding and simplifying, I got
$$
k^2 – k + 1 > 0.
$$
I can't solve this, so how would I work out the range of values that $k$ can be?
This was a textbook question, and the answer was
$$
\{k \text{ is a member of all real numbers such that $k$ does not equal 0}\}.
$$
I don't understand why this is the answer, can someone explain how we get there please?
Best Answer
You have done most of the work already. What's left is to note that, as a function of $k$, $k^2-k+1$ is a parabola with minimum value $3/4$ for $k=1/2$. In other words, $k^2-k+1 > 0$ for all $k$. However, we need to discard $k=0$, since, for that value, your original equation reads $2x+1=0$, which only has one root.