Q: Find the values of for which the quadratic expression 4^2 + 12 − is always positive.
What I’ve tried:
The discriminant is > 0 therefore b^2-4ac>0
a=4 b=12k c=-k
This resulted in: 12k^2+16k>0
I am not sure where to go next to solve for the value of ‘k’, any help would be greatly appreciated, thanks!
You can simplify a bit the computations using the reduced discriminant: $$\Delta'=(6k)^2+4k=4k(9k+1)$$ The quadratic polynomial $4x^2 + 12x − $ has a constant sign (that of the leading coefficient) if & only if $\Delta'<0$. As the leading coefficient of $\Delta'$ (as a quadratic polynomial in $k$) is positive, this means $k$ belongs to the interval of the roots, namely $\;\bigl(-\frac19,0\bigr)$.