Solving $ \sqrt{5x+1}+\sqrt{x-1}=2$

Question

How to solve:

$$ \sqrt{5x+1}+\sqrt{x-1}=2$$

I can tell that 1 is a solution but I am not sure how to solve this algebraically, do i start by squaring both sides?

Answer 1

Hint:

Since the expression involves $\sqrt{x-1}$, we must have $x \geq 1$. But in that case, $\sqrt{5x+1} \geq \sqrt6 > 2$. What does that imply?

Answer 2

The pedestrian approach would be$$\sqrt{5x+1}+\sqrt{x-1}=2\\(5x+1)+2\sqrt{5x+1}\sqrt{x-1}+(x-1)=4\\ 4(5x+1)(x-1)=(4-(5x+1)-(x-1))^2$$ and you have a quadratic
but for this case Théophile has a good answer

Answer 3

Hint:

As $x-1\ge0, 5x+1\ge?$

So what will be the minimum value of the sum?

Answer 4

$$ \sqrt{5x+1}+\sqrt{x-1}=2$$

Squarring both sides,we get

$$ 5x+1+x-1+2\sqrt{5x+1}\sqrt{x-1}=4$$ or$$ 6x+2\sqrt{5x+1}\sqrt{x-1}=4$$ or$$ 2\sqrt{5x+1}\sqrt{x-1}=4-6x$$ or$$ \sqrt{5x+1}\sqrt{x-1}=2-3x$$

Again squarring both sides,we get

$$ (5x+1)(x-1)=(2-3x)^2$$ or$$ 5x^2-4x-1=4+9x^2-12x$$ or$$ 4x^2-8x+5=0$$

Hence the roots of the equation are: $$x=\frac{8\pm\sqrt{64-4\cdot 5\cdot 4}}{2\cdot 4} = \frac{8\pm\sqrt{-16}}{8}$$ or$$x=\frac{8\pm4i}{8}=1\pm\frac{i}{2}$$

Answer 5

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: $$ (5x + 1) + 2\sqrt{(5x^2-4x-1)} + (x-1) = 4 $$

And thus: $$ \sqrt{(5x^2-4x-1)} = 2 - 3x $$

Now squaring both sides again: $$ 5x^2-4x-1 = 4 - 12x +9x^2 $$

Which gives the quadratic equation:

$$4x^2 -8x +5 = 0$$

Solving the quadratic equation gives two solutions: $x=1+i/2$ and $x=1-i/2$. So no real solutions exist.