How I see it:
$$(\sqrt{2-x})^2 = x^2 $$
$$2-x = x^2 \implies x^2 + x - 2 = 0$$
$$x^2 + x - 2 = (x+2)(x-1)$$
So the solutions for $x$ are $-2$ or $1$, but my textbook says $1$ is the only answer.
If I solve for $x=-2$ then I get $\sqrt{2-(-2)}=-2 \implies \pm2 = -2$ so isn't it true that $-2$ or $2$ equals $-2$? Shouldn't $-2$ also be a solution?
On
$$\sqrt{2-x} = x \implies 2-x = x^2$$ by squaring both sides, so after solving this quadratic, we'll need to check other solutions in case we introduced any spurious solutions. Anyway, on we go! $$2-x = x^2 \iff (x+2)(x-1) = 0$$
So $x=-2$ or $x=1$ satisfies the quadratic. Check both solutions in the original equation and throw away the one that does not work.
Basically, the spurious solution problem arises since $a=b \implies a^2 = b^2$ but this is only a one-way implication, the reverse isn't true. So you cannot say $a = b \iff a^2 = b^2$ because that just isn't true. What if $b=-a$?
On
No, $-2$ should not be a solution. The square root of a positive number is by definition always positive, meaning that $\sqrt{4}=2$ and not $\sqrt{4}=\pm 2$. Thus you only get $2=-2$ which is obviously not true.
Another way to look at it: squaring both sides of an equation does not need to be an equivalent transformation. An easy example to consider would be $$1=x \iff (1)^2=(x)^2 \iff \pm1=x,$$ so from $x=1$ we somehow get two solutions $x=\pm 1$.
So whenever you square both sides of an equation, you might get false solutions. To avoid this, you either check every solution you get in the original equation, or you first determine when LHS/RHS of the equation are both positive.
On
Hint:
since $\sqrt{2-x}\ge 0$ (by definition of square root) and it is real-defined only if $2-x\ge 0$ we have:
$$ \sqrt{2-x}=x \iff \begin{cases} 2-x \ge 0\\ x \ge 0\\ 2-x=x^2 \end{cases} $$
in general: the solution of an equation of the form $\sqrt{A}=B$ is equivalent to the system: $$ \begin{cases} A\ge 0\\ B \ge 0\\ A=B^2 \end{cases} $$
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What this equation is saying is that a function on the left hand side is the same as the function on the right hand side. And you are trying to find the value of x that makes this statement true. That being said your explanation is correct if you consider each side of these equations as IMPLICIT equations. The image above shows the function "x" in blue, and the function "$\sqrt{2-x}$" in red. If you consider "$\sqrt{2-x}$" as an implicit equation, instead of a function you get the green and read curves together, and where they intersect with the blue function, those are your solutions.
On
The important concepts here are equivalence and implications. The operation to square both sides is not an equivalence relation $\Leftrightarrow$. This means that the sets of solutions do not need to be the same before as after.
Squaring is a forward implication $\Rightarrow$ which means that all solutions will be preserved - we are sure to not lose any solution on the way.
But it is not a backward implication $\Leftarrow$, which would mean that the solution set on the right would be preserved when going to the left.
So all in all we are sure that after squaring all solutions to the original equation will be preserved, but that there may also be introduced other, new solutions.
This is a common mistake that a lot of people fall into. When they see $\sqrt{9}$ for example, they might say "this is the number such that when you square it you get $9$" and so they get $\sqrt{9} = \pm 3$. But this is wrong.
When you see the square root of something, the thing you should say to yourself is "this is the positive number such that when you square it, you get the other number". So $\sqrt{81}$ is the positive number such that when you square it, you get $81$. So $\sqrt{81} = 9$.
Now, if you are asking for the solutions of $x^{2} = 81$, then this would be $x = \pm \sqrt{81} = \pm 9$, but the square root of a number is always positive. So you are asking yourself the question to solve $x^{2} = 81$, but you are asking it for when trying to find $\sqrt{81}$, and that is where you are going wrong.