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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?

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  • $\begingroup$ I would like to see how you arrived at $x^2+6x+2$ because then we can see where you went wrong... $\endgroup$ – imranfat Mar 9 '16 at 16:47
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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 $t=0$, $x=1$.

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  • $\begingroup$ neat solution :) $\endgroup$ – Imago Mar 9 '16 at 17:17
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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.

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  • $\begingroup$ 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. $\endgroup$ – Imago Mar 9 '16 at 17:11
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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.

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