Finding for the value of constant $k$ in an inequality for unique solution I got this question from an old math questionnaire and got stuck. 
For which value of the constant $k$ will the following inequality have a unique solution?
$$
9k^2 (x-5)^2 - 125k^2 \geq (9+5k^2)(x^2 - 10x) + 225
$$
I tried simplifying the inequality. I expressed it as a quadratic inequality and set the discriminant as $0$ so that I would have a unique solution. But as I was about to solve for $k$, the discriminant cancelled to $0$ and I could not solve the value of $k$ anymore.
I am also new here in math.stackexchange and I do not know how to properly encode equations and inequalities. So please, bear with me. 
 A: Put $u = (x-5)^2 \implies u \ge 0$, and the inequality reduces to $f(u) = (4k^2-9)u + 100k^2 \ge 0$ for all $u \ge 0$. Observe $f$ is a linear function in $u$, thus we have: $4k^2-9 \ge 0$ or $|k| \ge \dfrac{3}{2}$.
A: See, your inequality is as follows,
\begin{align*}
&\quad\quad\ \  9k^2 (x-5)^2 - 25k^2-(9+5k^2)(x^2 - 10x) - 225 \ge 0\\
& \implies 9k^2(x^2-10x+25)-25k^2-9x^2+90x-5x^2k^2+50xk^2+225 \ge 0\\
& \implies9k^2x^2-90k^2x+225k^2-25k^2-9x^2+90x-5k^2x^2+50xk^2+225 \ge 0\\
& \implies x^2(4k^2-9)-x(40k^2-90)+200k^2+225 \ge 0\\
\end{align*}
This is a quadratic equation in $x$ and it is always greater than $0$  so, the discriminant has to be less than or equal to zero. Hence, $$ (40k^2-90)^2-4(4k^2-9)(200k^2+225)\le 0 .$$ Solve this for $k$ and get the solution.
I think I got your doubt, if not tell me.
A: You want that $$9k^2 (x-5)^2 - 125k^2 \geq (9+5k^2)(x^2 - 10x) + 225$$ that is to say $$9k^2 (x-5)^2 - 125k^2 - (9+5k^2)(x^2 - 10x) - 225\geq 0$$ Expand and group terms to get $$\left(4 k^2-9\right) x^2+\left(90-40 k^2\right) x+(100 k^2-225) \geq 0$$ Rearranging, you have $$\left(4 k^2-9\right) x^2-10 \left(4 k^2-9\right)x+25\left(4 k^2-9\right) \geq 0$$ that is to say  $$\left(4 k^2-9\right)(x^2-10x+25) \geq 0\implies \left(4 k^2-9\right)(x-5)^2\geq 0$$ I am sure that you can take it from here
