Solving for all values of k if a quadratic equation containing it has two distinct real roots

Question

The equation $$x^2-2kx+1=0$$ has two distinct real roots. Find the set of all possible values of k.

What I did:

1. $$b-4ac=0$$
2. $$(-2k)^2-4=0$$
3. $$(-2k)^2=4$$
4. $$-2k=2$$
5. $$k=(-1)$$

From this, I'm unsure where to go-- it says find the set. I understand that $$k=1$$ as well as -1 but how is that a set? Like I think the question wants me to give my answer in set notation but I'm unsure how to do that.

Answer 1

solving the given equation we obtain $$x=k\pm\sqrt{k^2-1}$$ thus we obtain two real solutions if $$k^2>1$$ or $$(k-1)(k+1)>0$$ can you proceed?

Answer 2

The values of $k$ are precisely the solutions of $\Delta>0$ that's $k^2-1>0$, which gives $k<-1\vee k>1$.

Answer 3

For distinct real roots, $$b^2-4ac\gt0$$ $$(-2k)^2-4\gt0$$ $$k^2-1\gt0$$ $$(k+1)(k-1)\gt0$$ $$k\in(-\infty,-1)\cup(1,\infty)$$

Answer 4

The quadratic formula uses $\sqrt{b^2-4ac}$. For there to be $2$ distinct real roots, $b^2-4ac\gt 0$, otherwise if $b^2-4ac=0$ the two roots are real but the same, and if $b^2-4ac\lt 0$ both roots are complex.

So we need to find the set of solution to $b^2-4ac\gt 0$, which, using the equation given, is solutions to $(-2k)^2-4\gt0$, which simplifies to $k^2>1$ or $|k|>1$.

So the set $S$ of solutions is $S=\{k: |k|\gt1\}$.