Find x for $\sqrt{(5x-1)}+\sqrt{(x-1)}=2$ Solve:
$$\sqrt{(5x-1)}+\sqrt{(x-1)}=2$$
When $x=1$, we get the following equation to equal to $2$
I've been trying to solve this problem but when I square both sides and simplify I end up with:
$$x^2+6x+2=0$$ and of course $x=1$ cannot be a solution. So im not sure what im doing wrong. Any help on this problem?
 A: Let $\sqrt{x-1}=t$. The equation becomes
$$\sqrt{5t^2+4}=2-t$$or by squaring and regrouping
$$4t^2+4t=0.$$
As $t$ is positive, the only solution is $t=0$, $x=1$.
A: Yes, squaring both sides gives:
$$ (5x - 1) + 2\sqrt{5x-1}\sqrt{x-1} + (x-1) = 4 $$
Then: 
$$ (5x - 1) + 2\sqrt{(5x-1)(x-1)} + (x-1) = 4 $$
Which simplifies to:
$$ (5x - 1) + 2\sqrt{(5x^2-6x+1)} + (x-1) = 4 $$
And thus:
$$ \sqrt{(5x^2-6x+1)}  = 3 - 3x $$
Now squaring both sides again:
$$ 5x^2-6x+1  = 9- 18x +9x^2 $$
Which gives the quadratic equation:
$$4x^2 -12x +8 = 0$$ 
And thus:
$$4(x^2 -3x +2) = 0$$ 
The potential solutions are thus the roots of $x^2 -3x +2$, i.e. $x=1$ or $x=2$. Plugging these values into the original equation we see that only $x=1$ is a solution. 
A: Perhaps a slightly tidier solution:
$$\begin{align}
\sqrt{5x-1}+\sqrt{x-1}&=2 \\
\sqrt{5x-1} &= 2-\sqrt{x-1} &\text{from regrouping}\\
5x-1&=4-4\sqrt{x-1}+(x-1) & \text{after squaring}\\
4x-4&=4\sqrt{x-1} &\text{after regrouping again}\\
(x-1)^2&=x-1 &\text{divide by 4 and square again}\\
(x-1)^2-(x-1)&=0 \\
(x-1)(x)&=0 &\text{factor}
\end{align}$$
so either $x-1=0$ or $x=0$. Plugging back into the original equation we see that $x=1$ is a solution and $x=0$ is not.
