Determining the value of k in f(x)

Question

The question I am struggling with is this:

Determine the value of $k$ if $f(x)=(k-2)x^2+6x+1$ has exactly one real zero

I have tried a couple of things, but I keep ending up with no solution.

Answer 1

If This quadratic equation has only one real root, then it must have $b^2-4ac=0$

Using that,$$36-4\times(k-2)=0$$ $$k=11$$

Answer 2

If you use the quadratic equation to expand $f(x)$ you get:

$f(x)=(k-2)x^2+6x+1=0 \implies x=-6\pm\frac{\sqrt{36-4(k-2)1}}{2(k-2)}=-6\pm\frac{\sqrt{36-4k+8}}{2k-4}=$ $-6\pm\frac{\sqrt{44-4k}}{2k-4}=-6\pm\frac{44-4k}{(2k-4)^2}$

Now, the problem is telling you explicitely f(x) has only 1 real root, that means to meet that condition the second term of the equation $\pm\frac{44-4k}{(2k-4)^2} = 0$, so you get $44-4k=0\implies k=\frac{44}{4}=11$