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!
 A: Denote $t:=\sqrt{45x^2-30x+1}$. Then we observe that
$$t^2=45x^2-30x+1=-5(7+6x-9x^2)+36.$$
As a result, it follows that $$t=-\frac{t^2-36}{5}\implies t^2+5t-36=0\implies(t+9)(t-4)=0.$$
Since $t\geq 0$, it follows that $t=4$. Therefore, $$45x^2-30x+1=16\implies45x^2-30x-15=0\implies 3x^2-2x-1=0.$$
Hence $(3x+1)(x-1)=0$. Either $x=1$ or $x=-1/3$.
A: Let $u=9x^2-6x$ and rewrite the equation as
$$\sqrt{5u+1}=7-u$$
Squaring both sides gives $5u+1=49-14u+u^2$, or
$$u^2-19u+48=(u-16)(u-3)=0$$
We see that $u=16$ is not a solution, since $\sqrt{81}\not=-9$. This leaves us with $u=3$, which is a valid solution, since $\sqrt{16}=4$, and from this we have
$$9x^2-6x=3\implies3x^2-2x-1=(x-1)(3x+1)=0$$
which gives $x=1$ and $x=-1/3$ as the complete solution set.
Remark: What makes this work so nicely is that $45:30=9:6$. If the coefficients of $x^2$ and $x$ on the two sides hadn't been in proportion, the solution would have been much more involved.
A: $$45x^2-30x+1=(-9x^2+6x+7)^2$$
$$-5(-9x^2+6x+7)+36=(-9x^2+6x+7)^2$$
Substitute $u=-9x^2+6x+7$ and solve the quadratic equation.
A: 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 $x = -\frac{1}{3}$.
