Show that the quadratic function $g(x)= 2x^2 - 4kx + k^2$ intersects the $x$-axis for all values of $k$.

Question

Should I use $b^2 - 4ac$? Is the answer $8k^2 > 0$?

Answer 1

You'll have two real solutions when $b^2-4ac>0$, like you found. But, you can still have one real solution when $b^2-4ac=0$.

So if $b^2-4ac \ge 0$ for all values of $k$, then, you'll have at least one real solution for any value of $k$.

In this case, you have $b^2-4ac = 8k^2$, and since $8>0$ and $k^2 \ge 0$ (no matter what value of $k$ you plug in), you've shown that $b^2-4ac \ge 0$ for all values of $k$, and you're done!

Answer 2

HINT

What about completing squares?

\begin{align*} 2x^{2} - 4kx + k^{2} & = 2(x^{2} - 2kx) + k^{2}\\\\ & = 2(x^{2} - 2kx + k^{2} - k^{2}) + k^{2}\\\\ & = 2(x-k)^{2} - k^{2} \end{align*}

Can you take it from here?