algebra-precalculusquadraticsradicalssolution-verification
I'm trying to solve this equation:
$$2-x=-\sqrt{x}$$
Multiply by $(-1)$:
$$\sqrt{x}=x-2$$
power of $2$:
$$x=\left(x-2\right)^2$$
then:
$$x^2-5x+4=0$$
and that means:
$$x=1, x=4$$
But $x=1$ is not a correct solution to the original equation.
Why have I got it? I've never got a wrong solution to an equation before. What is so special here?
From
$$\begin{cases} x\sqrt{y} + y\sqrt{x} = 30 & \ \ (a)\\ x\sqrt{x} + y\sqrt{y} = 35& \ \ (b)\end{cases}$$
$3 \times$(a) + (b) gives :
$$(\sqrt{x}+\sqrt{y})^3=5^3\ \ \iff \ \ \sqrt{x}+\sqrt{y}=5 \ \tag{1}$$
Besides, (b) - (a) gives :
$$x(\sqrt{x}-\sqrt{y})-y(\sqrt{x}-\sqrt{y})=5 \ \iff$$
$$(\sqrt{x}-\sqrt{y})^2(\sqrt{x}+\sqrt{y})=5 \tag{2}$$
Taking (1) into account in (2), on gets :
$$\sqrt{x}-\sqrt{y}=\pm1\tag{3}$$
Gathering (1) and (3), we obtain easily the two solutions :
$$(x,y)=(4,9) \ \text{and} \ (x,y)=(9,4) $$
as confirmed by the following graphical representation ((a) in blue and (b) in red) :
I'll sum up the concepts it seems a few have addressed in the comments.
Setting the square root of something equal to a negative will only result in an imaginary solution, as I'm sure you are aware. Thus, squaring both sides will ALWAYS result in an extraneous solution. So, if you can show that an equation is of the form "square root = a negative," then you know the equation has no real solutions. So yes, your logic is correct.
Good of you to be thinking about equations such as these in a more general sense!
Best Answer
This is because the equation $\;\sqrt x=x-2$ is not equivalent to $x=(x-2)^2$, but to $$x=(x-2)^2\quad\textbf{and}\quad x\ge 2.$$ Remember $\sqrt x$, when it is defined, denotes the non-negative square root of $x$, hence in the present case, $x-2 \ge 0$, i.e. $x$ must be at least $2$.