# 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?

• I would like to see how you arrived at $x^2+6x+2$ because then we can see where you went wrong... – imranfat Mar 9 '16 at 16:47

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 $t=0$, $x=1$.

• neat solution :) – Imago Mar 9 '16 at 17:17

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: $$5x^2-6x+1 = 9- 18x +9x^2$$

$$4x^2 -12x +8 = 0$$
$$4(x^2 -3x +2) = 0$$
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.
• One may alternatively use: $f(x) := \sqrt(5x-1) + \sqrt(x-1) - 2$ is monotonously increasing, defined on $[1, \infty)$ and $f(1) = 0$ (I prefer this type of argument, since I am usually pretty unrelieable, if I have to compute something. – Imago Mar 9 '16 at 17:11
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.