Find the range of values of k

Question

Find the range of values of $k$ for which the equation

$$x^2-2kx+k^2-2k=6$$

has real roots. Find the roots in terms of $k$.

Answer 1

Hint: You can solve quadratic equations? You should consider $k$ as a parameter, (think of it like any number). Also you could substitute $k^2-2k \to q$ and solve for $x$, if the constant term $k^2-2k$ bears problems. So if you know how to solve quadratic equations of the form $$x^2 + px + q$$ then this is nothing else: $p = -2k$ and $q=k^2-2k$

Secondly, the solution of the equation is:

$$x_{1/2} = -\frac{p}{2} \pm \sqrt{(\frac{p}{2})^2 -q}$$

Answer 2

$$x^2-2kx+k^2-2k-6 = (x-k)^2-2k-6.$$Hence the equation has real roots iff $$ -2k-6\le 0\iff k\ge -3. $$