Solving $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $. I have the equation $ \sqrt{x - 4} + \sqrt{x - 7} = 1 $.
I tried to square both sides, but then I got a more difficult equation:
$$
2 x - 11 + 2 \sqrt{x^{2} - 11 x - 28} = 1.
$$
Can someone tell me what I should do next?
 A: Suppose that $x$ is a solution. Flip things over. We get
$$\frac{1}{\sqrt{x-4}+\sqrt{x-7}}=1.$$
Rationalizing the denominator, and minor manipulation gives
$$\sqrt{x-4}-\sqrt{x-7}=3.$$
From this, addition gives $2\sqrt{x-4}=4$. The only possibility is $x=8$, which is not a solution.  
A: squaring gives
$$x-4+x-7+2\sqrt{x-4}\sqrt{x-7}=1$$
$$\sqrt{x-4}\sqrt{x-7}=12-2x$$
$$(x-4)(x-7)=36+x^2-12x$$
$$x=8$$
but $8$ fulfills not our equation.
A: There is no real solution since you need $x \geq 7$ for sure. The left hand side is an increasing function for $x \geq 7$ and so it's minimum is $\sqrt{3}>1$. 
A: $$\sqrt{x-4}+\sqrt{x-7}=1<=>$$
$$-11+2\sqrt{(x-7)(x-4)}=12-2x<=>$$
$$4(x-7)(x-4)=(12-2x)^2<=>$$
$$4x^2-44x+112=4x^2-48x+144<=>$$
$$4x-32=0<=>$$
$$4(x-8)=0<=>$$
$$x-8=0<=>$$
$$x=8$$
BUT THIS SOLUTION IS INCORRECT SO THERE ARE NO SOLUTIONS!!!!!
A: We cannot assume that a root/solution always exists, especially when sqrt is present. At first test with a few values of $x$ to ascertain whether  it can be satisfied approximately. The squaring sometimes introduces possibility of the second sign before the radical.
The equation as you gave ( positive value for both sqrts) has no real solution as x must be > 7.
However, $$ -\sqrt{x-7} + \sqrt{x-4} = 1 $$ has the solution $ x = 8. $
