For what $k$ is $f(x) = kx^2-2x+k$ negative for all values of $x$?

Question

What are the values of $k$ for which the quadratic function $f(x) = kx^2-2x+k$ is negative for all values of $x$?

The values of $k$ should definitely be negative.

Answer 1

First off, we require $k<0$ then we require that $f$ has no roots so that it is always negative. That is, we need its discriminant to be $< 0$. i.e: $$4 - 4k^2 < 0 \iff 1- k^2 < 0 \iff k^2 > 1$$ since we know that $k<0$ then this becomes $k < -1$.

Answer 2

A different, although perhaps slightly more complicated, approach than the other answers:

Find the vertex of this parabola and determine which values of $k$ put the vertex below the $x$-axis. Combine that with the fact you've already found: $k < 0$. Completing the square is one method of finding the vertex:

$$ kx^2 - 2x + k = k\left(x^2 - \frac{2}{k}x\right) + k = k\left(x - \frac{1}{k}\right)^2 + k - \frac{1}{k} $$

The $y$-coordinate of the vertex is $k - \frac{1}{k}$, and we want this to be negative, so...

Answer 3

Hint: complete the square.

$f(x) = kx^2-2x+k = k(x^2-\frac{2}{k}x+1) = k(x-\frac{1}{k})^2 + k - \frac{1}{k}$. As you noted, $k$ should be negative. In that case, $k(x-\frac{1}{k})^2 \le 0$ for all $x$, with equality at $x=1/k$. So, we simply need $k-1/k < 0$, which for negative $k$ is the same as $k^2 > 1$. So $k<-1$.