Why am I getting extraneous solutions solving $\frac{1+\sqrt{1-x}}{x-\sqrt{1-x^{2}}}=2 x$?

Question

I came across a beautiful problem:

Solve $$\frac{1+\sqrt{1-x}}{x-\sqrt{1-x^{2}}}=2 x$$

My approach:

Obviously $x=0$ is not a solution. Now we have:

$$1+\sqrt{1-x}=2x^2-2x\sqrt{1-x^2} \tag1$$

Squaring the above equation both sides we get $$2(1+\sqrt{1-x})-x=4x^4+4x^2(1-x^2)-8x^3\sqrt{1-x^2} \tag2$$

From $(1)$ we get $$2(2x^2-2x\sqrt{1-x^2})-x=4x^2-8x^3\sqrt{1-x^2} \tag3$$

$\implies$

$$(8x^3-4x)\sqrt{1-x^2}=x \tag4$$

So we have: $$(8x^2-4)\sqrt{1-x^2}=1 \tag5$$

Letting $\sqrt{1-x^2}=p, p>0$ We get $$8p^3-4p+1=0 \tag6$$ which gives $p = 0.5, \frac{\sqrt{5}-1}{4}$

So using $x=\pm \sqrt{1-p^2}$ we get four values of $x$ as: $$x=\frac{\pm \sqrt{3}}{2}, \frac{\pm \sqrt{10+2 \sqrt{5}}}{4} \tag7$$ Out of which only $$x=\frac{-\sqrt{3}}{2}, \frac{-\sqrt{10+2 \sqrt{5}}}{4}$$ Will satisfy.

But at which step the positive extraneous roots boiled down and how to get rid them in this approach?

Answer 1

we get four values of $x:$ $$x=\frac{\pm \sqrt{3}}{2}, \frac{\pm \sqrt{10+2 \sqrt{5}}}{4} \tag7$$ out of which only $$x=\frac{-\sqrt{3}}{2}, \frac{-\sqrt{10+2 \sqrt{5}}}{4}$$ will satisfy.

Actually, among these four solutions, only $x=\frac{\sqrt{3}}{2}$ is extraneous; it was introduced, together with the other extraneous solution $x=0,$ in step $(3)$. Every other step—including the two squaring steps—turns out to be “reversible” (preserve the solution set), even as you have not justified how (for example, by specifying a relevant implicit constraint).

$$\frac{1+\sqrt{1-x}}{x-\sqrt{1-x^{2}}}=2 x$$

$$1+\sqrt{1-x}=2x^2-2x\sqrt{1-x^2} \tag1$$

Squaring the above equation, we get $$2(1+\sqrt{1-x})-x=4x^4+4x^2(1-x^2)-8x^3\sqrt{1-x^2} \tag2$$

From $(1),$ we get $$2(2x^2-2x\sqrt{1-x^2})-x=4x^2-8x^3\sqrt{1-x^2}\tag3$$

Substituting equation $(1)$ into equation $(2)$ to create equation $(3)$ has expanded the candidate solution set, as $(1)$ is a conditional equation rather than an identity (like $(x+3)^2=x^2+6x+9,$ which is universally true).

Since equation $(1)$ holds at our required solution points, they will not be discarded. However, at every other point, equation $(1)$ is false, so deductive explosion will occur, so a spurious conclusion (extraneous solution) may ensue even as our subsequent argument remains valid.

Here's a silly clearer example to illustrate. We want to solve the equation $$x^2+1=5.\tag i$$ Thus, $$x^2=4.\tag {ii}$$ Substituting $(\mathrm{ii})$ into $(\mathrm i):$ $$5=5\\x+5=x+5\\x=x.$$ Therefore, every number is a potential/candidate solution of equation $(\mathrm i).$

To be clear, neither of the above substitution steps is invalid: they do not discard actual solutions, and merely turn out to have created extraneous solutions which then need to be sifted out.