Proving that $\sqrt{x-1}=3+\sqrt{x}$ has no real solutions Prove that there is no real number $x$ such that $\sqrt{x-11}=3+\sqrt{x}.$
The book suggested a hint "Start Squaring by both sides". Now I have tried this as. Squaring both sides we get $x-1=9+6\sqrt{x}+x.$ It follows that $0=10+6\sqrt{x}$, so $\sqrt{x}=-10/6. \tag3$
Now I do not know what I do next.
 A: Since you always have $\sqrt x\geqslant0$, the equation$$10+6\sqrt x=0\tag1$$has no solutions. But every solution of the equation$$\sqrt{x-1}=3+\sqrt x\tag2$$is also a solution of $(1)$. Therefore, $(2)$ also has no solutions.
A: I think José Carlos Santos's answer is best, but it never hurts to see an alternative approach. This one makes use of the difference of two squares:
$$\sqrt{x-1}=3+\sqrt{x}\qquad (1)$$
$$\implies\sqrt{x-1}-\sqrt{x}=3$$
$$\implies\left(\sqrt{x-1}-\sqrt{x}\right)\left(\sqrt{x-1}+\sqrt{x}\right)=3\left(\sqrt{x-1}+\sqrt{x}\right)$$
$$\implies -1=3\left(\sqrt{x-1}+\sqrt{x}\right)\qquad (2)$$
but the right-hand side of $(2)$ is non-negative; therefore no real value of $x$ solves $(1).$
A: You could use proof by contradiction. Suppose, for the sake of contradiction, that there is a number $x$ such that $\sqrt{x-1}=3+\sqrt{x}$. Then, it follows from your manipulations that $\sqrt{x}=-10/6$. This is absurd, as $\sqrt{x}$ denotes the nonnegative square root of $x$, and so we can't have $\sqrt{x} <0$. Therefore, our original assumption, that there is a number $x$ such that $\sqrt{x-1}=3+\sqrt{x}$, must be false; that is,$\sqrt{x-1}=3+\sqrt{x}$ has no solutions.
A: Alternative thinking:
We know that,
$$\sqrt{x-1} < \sqrt{x}$$
which implies that,
$$3 + \sqrt{x-1} < 3 + \sqrt{x}$$
and hence,
$$\sqrt{x-1} \not = 3 + \sqrt{x}$$
as the right-hand side is always greater than the left-hand side.
A: The range of $\sqrt{x}$ (the principle square root) does not include negative numbers because it is defined to be a function. Therefore, the equation $\sqrt{x-1} = 3+\sqrt{x}$ has no real solutions because you get $\sqrt{x} = -\dfrac{10}{6}$ when isolating the $x$ variable.
