\documentclass[a4paper]{article}
\usepackage{amsmath}
\begin{document}
\title{Don't Square Twice}
\author{Amit Yaron}
\date{Jan 25, 2025}
\maketitle
\section{Introduction}
I like seeing thumbnails of math videos on YouTube, but without watching
the video. Instead, I prefer making a document of the solution. One example is:
\[
  \sqrt{x-x^3}+\sqrt{x^2-x^3}=1
\]
One way that leads to a quartic equation is to subtract one of the sqare
roots to the right-hand side, and square both sides. Let us try something
else.
\section{Substitution}
Let
\[
  a=\sqrt{x-x^3},\ b=\sqrt{x^2-x^3}
\]
Then, we can see that:
\[
\begin{cases}
  a+b=1\\
  a^2-b^2=x-x^2
\end{cases}
\]
By dividing the second equation by the first one, we get:
\[
  a-b=x-x^2
\]
Now,
\[
  a+b=1\Rightarrow b=1-a\Rightarrow a-b=1-2b=x-x^2=\frac{b^2}x
\]
So, we can plug $x=\frac{b^2}{1-2b}$ into $b=\sqrt{x^2-x^3}$, and solve:
\begin{equation}
  \sqrt{\frac{b^4}{(1-2b)^2}-\frac{b^6}{(1-2b)^3}}=b
\end{equation}
Now, is $b=0$ a solution?
\begin{align*}
  &b=0\Rightarrow\\
   \Rightarrow &x=\frac{b}{1-2b}=\frac{0}1=0\Rightarrow\\
   \Rightarrow &\sqrt{x-x^3}+\sqrt{x^2-x^3}=0\neq1
\end{align*}
which contradicts the definition of our equation. So, we can divide both sides
of equation (1) by $b$, or the radicant by $b^2$
\begin{align*}
  &\sqrt{\frac{b^2}{(1-2b)^2}-\frac{b^4}{(1-2b)^3}}=1\Rightarrow\\
  \Rightarrow &\frac{b^2}{(1-2b)^2}-\frac{b^4}{(1-2b)^3}=1\Rightarrow\\
  \Rightarrow &(1-2b)b^2-b^4=(1-2b)^3\Rightarrow\\
  \Rightarrow &b^2-2b^3-b^4=1-6b+12b^2-8b^3\Rightarrow\\
  \Rightarrow &b^4-6b^3+11b^2-6b+1=0\Rightarrow\\
  \Rightarrow &(b^4+1)-6(b^3+b)+11b^2=0\Rightarrow\\
  \Rightarrow &(b^2+\frac1{b^2})-6\big(b+\frac1b\big)+11=0\Rightarrow\\
  \Rightarrow &(b^2+\frac1{b^2}+2b\big(\frac1b)-2b\big(\frac1b)-6\big(b+\frac1b\big)+11=0\Rightarrow\\
  \Rightarrow &\big(b+\frac1b)^2-2-6\big(b+\frac1b\big)+11=0\Rightarrow\\
  \Rightarrow &\big(b+\frac1b)^2-6\big(b+\frac1b\big)+9=0\Rightarrow\\
  \Rightarrow &[\big(b+\frac1b\big)-3]^2=0\Rightarrow\\
  \Rightarrow &b+\frac1b-3=0\Rightarrow\\
  \Rightarrow &b^2+1-3b=0\Rightarrow\\
  \Rightarrow &b^2-3b+1=0\Rightarrow\\
  \Rightarrow &b=\frac{3\pm\sqrt{(-3)^2-4\cdot1\cdot1}}2=\frac{3\pm\sqrt5}2
\end{align*}
\section{Dropping the Extraneous Root}
Let us refer back to the substitution at the beginning of Section 2:\\
$a$ and $b$ are square root, thus they cannot be negative.\\
and $a+b=1\Rightarrow a=1-b$
If we take:
\[
  b=\frac{3+\sqrt{5}}2
\]
then:
\[
  a=1-\frac{3+\sqrt5}2=\frac{2-(3+\sqrt5)}2=\frac{-1-\sqrt5}2<0
\] 
But, $a$ cannot be negative, thus our solution is
\[
  b=\frac{3-\sqrt5}2
\]
\section{Back Substitution}
\[
  x = \frac{b^2}{1-2b}
\]
Let us square $b$:
\begin{align*}
  b^2&=\big(\frac{3-\sqrt5}2\big)^2\\
  &=\frac{3^2+\sqrt5^2-2\cdot3\sqrt5}4\\
  &=\frac{14-6\sqrt5}4\\
  &=\frac{7-3\sqrt5}2
\end{align*}
And what is $1-2b$?
\[
  1-2b=1-2\cdot\frac{3-\sqrt5}2=1-(\frac3-\sqrt5)=\sqrt5-2
\]
Thus,
\begin{align*}
  x&=\frac{b^2}{1-2b}\\
  &=\frac{7-3\sqrt5}{2\sqrt5-4}\\
  &=\frac{(7-3\sqrt5)(2\sqrt5+4)}{(2\sqrt5-4)(2\sqrt5+4)}\\
  &=\frac{14\sqrt5-30+28-12\sqrt5}{20-16}\\
  &=\frac{2\sqrt5-2}4\\
  &=\frac{\sqrt5-1}2
\end{align*}
To test the solution:
\begin{align*}
  &x=\frac{\sqrt5-1}2\Rightarrow\\
  \Rightarrow &x^2=\frac{5+1-2\sqrt5}4\Rightarrow\\
  \Rightarrow &x^2=\frac{3-\sqrt5}2\Rightarrow\\
  \Rightarrow &x^2=\frac{2+1-\sqrt5}2\Rightarrow\\
  \Rightarrow &x^2=1+\frac{1-\sqrt5}2\Rightarrow\\
  \Rightarrow &x^2=1-x\Rightarrow\\
  \Rightarrow &x^3=x(1-x)\Rightarrow\\
  \Rightarrow &x^3=x-x^2\Rightarrow\\
  \Rightarrow &x^3=x-(1-x)\Rightarrow\\
  \Rightarrow &x^3=2x-1 
\end{align*}
Thus,
\[
  x-x^3=x-(2x-1)=1-x=x^2\Rightarrow \sqrt{x-x^3}=x
\]
And,
\begin{align*}
  x^2-x^3&=(1-x)-(2x-1)\\
  &=2-3x\\
  &=2-\frac{3\sqrt5-3}2\\
  &=\frac{7-3\sqrt5}2\\
  &=\frac{14-6\sqrt5}4\\
  &=\frac{9+5-6\sqrt5}4\\
  &=\frac{3^2+\sqrt5^2-2\cdot3\sqrt5}4\\
  &=\frac{(3-\sqrt5)^2}4\\
  &=\big(\frac{3-\sqrt5}2\big)^2\\
  &=(1-x)^2
\end{align*}
Thus,
\[
  \sqrt{x^2-x^3}=1-x\Rightarrow \sqrt{x-x^3}+\sqrt{x^2-x^3}=1
\]
\end{document}