<s>[INST] Solve the equation ${\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}} = 9x-3$ I tried factoring the expression inside the square root, but that does not seem to help. Squaring the equation makes it even more terrible.
Can anyone provide a hint about what should be done? 
 [/INST] Rewrite your equation so that there is only one square root on each side. We get
$$\sqrt{4x^2+5x+1}=2\sqrt{x^2-x+1}+9x-3.$$
Squaring both sides we get
$$
\begin{align}
4x^2+5x+1&=4(x^2-x+1)+4(9x-3)\sqrt{x^2-x+1}+(81x^2-54x+9)\\
&=85x^2-58x+13+4(9x-3)\sqrt{x^2-x+1}\end{align}.$$
We get
$$81x^2-63x+12=-4(9x-3)\sqrt{x^2-x+1}.$$
Dividing by $3$, we get
$$27x^2-21x+4=-4(3x-1)\sqrt{x^2-x+1}.$$
Apply factoring at the LHS, we get
$$(9x-4)(3x-1)=-(3x-1)\cdot 4\sqrt{x^2-x+1}$$
Hence,
$$(9x-4)(3x-1)+(3x-1)\cdot 4\sqrt{x^2-x+1}=0.$$
Thus, 
$$(3x-1)\cdot\Big[9x-4)+ 4\sqrt{x^2-x+1}\Big]=0.$$
Either
$x=\frac{1}{3}$ or
$$9x-4=-4\sqrt{x^2-x+1}.$$
Squaring again to both sides, we get
$$81x^2-72x+16=16(x^2-x+1).$$
We get
$$65x^2-56x=0.$$
Either $x=0$ or $x=\frac{56}{65}.$ Only $x=\frac{1}{3}$ satisfies the original equation.
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