<s>[INST] Solve the equation $\sqrt{45x^2-30x+1}=7+6x-9x^2$ Solve the equation $$\sqrt{45x^2-30x+1}=7+6x-9x^2.$$
So we have $\sqrt{45x^2-30x+1}=7+6x-9x^2\iff \begin{cases}7+6x-9x^2\ge0\\45x^2-30x+1=(7+6x-9x^2)^2\end{cases}.$ The inequality gives $x\in\left[\dfrac{1-2\sqrt{2}}{3};\dfrac{1+2\sqrt{2}}{3}\right].$ I am not sure how to deal with the equation. Thank you in advance!
 [/INST] From $45x^2-30x+1=(7+6x-9x^2)^2$, you have that $$81x^{4}-108x^{3}-135x^{2}+114x+48 = 0$$
This factors as $$3(x-1)(3x+1)(9x^2-6x-16) = 0$$
So the roots of that polynomial are $x = 1, -\frac{1}{3}, \frac{1\pm\sqrt{17}}{3}$. Out of these, the only ones in the range $\left[\frac{1-2\sqrt{2}}{3} ,\frac{1+2\sqrt{2}}{3} \right]$ are $x = 1$ and $x = -\frac{1}{3}$.
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