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? 
 A: Notice that
$$\left({\sqrt{4x^2 + 5x+1}} - 2{\sqrt {x^2-x+1}}\right) \left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) = 9x-3.$$
Thus, 
$$9x-3 =  \left(9x-3\right)\left({\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}}\right) \implies $$
$$9x-3 = 0 \implies x=1/3$$ or
$${\sqrt{4x^2 + 5x+1}} + 2{\sqrt {x^2-x+1}} = 1.$$
But the minimal value of $x^2-x+1$ is $3/4$, which implies that 
$$ 2{\sqrt {x^2-x+1}} \geq \sqrt{3} > 1.$$
Therefore $x =1/3$ is the only real solution.
A: 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.
A: Try to multiply both sides by $(\sqrt{4x^2+5x+1}+2\sqrt{x^2-x+1})$. 
In the left side you will get $ 9x-3$ by difference of squares.
A: Move $2\sqrt {x^2 - x + 1}$ to the rhs, then square both sides.
