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$$\sqrt{x\sqrt{x} - x} = 1-x$$

I know the solution but have no idea how to solve it analytically.

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Did I convert the ASCII formula correctly? – Brian M. Scott May 9 '12 at 22:09
Yes, indeed you did. – Daan Michiels May 9 '12 at 22:10
You should probably say whether you want real or complex (or...) solutions. – Simon Markett May 9 '12 at 22:11
This is supposed to be a problem for students in 1st grade of high school. I'm sure they do not have a clue about how to solve a cubic equation yet... They have this problem among others where they use a substitution. How about using this one: sqrt(x) = t ? What is the simplest solution of the problem? – user1111261 May 10 '12 at 7:23
up vote 8 down vote accepted

If $\sqrt{x\sqrt{x}-x} = 1-x$, squaring both sides we have $x \sqrt{x} - x = (1-x)^2$, and then taking the $-x$ to the other side and squaring again we get $x^3 = (x + (1-x)^2)^2$. Simplify to $(x-1)(x^3 -2x^2 +x-1) = 0$, the second factor being irreducible over the rationals. Of course $x=1$ is a solution. The roots of the cubic are rather complicated. One is real (approximately $1.754877666$), but it is not a solution of the original equation because the right side would be negative and the square root of a positive number is positive. For the complex roots, you have to specify which branch of the square root you mean. If you mean the principal branch (i.e. nonnegative real part), the complex roots are also not solutions of the original equation.

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For the left hand side to be defined, you need $x\geq 1$ or $x=0$. Zero is not a solution. The left hand side will equal 0 for $x=1$ and will be strictly positive if $x>1$. For $x\geq 1$, the right hand side is equal to 0 if $x=1$ and will be strictly negative if $x>1$. This shows that the only real solution is $x=1$.

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Just a minor point: the left-hand side is defined for $x=0$ as well, but as it's clearly not a solution you can assume $x>0$ which then implies $x \geq 1$. – Antonio Vargas May 9 '12 at 22:17
Nice demonstration of astute observation making quick work of a problem! – rschwieb May 9 '12 at 22:19
@Antonio Vargas. You are right, I overlooked the case x=0. Edited. – Daan Michiels May 10 '12 at 7:30

The equation $$\sqrt{x\sqrt{x}-x} = 1-x$$ implies that both sides, and the argument of each square root, are all nonnegative. Thus any solution must obey $$\sqrt{x}\text{ exists}\implies x\ge0$$ $$0\le1-x\implies x\le1$$ $$0\le x\sqrt{x}-x=x\left(\sqrt{x}-1\right) \implies \sqrt{x}\ge1 \implies x\ge1$$ But then $1\le x\le 1 \implies x=1$ is the only solution.

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Just writing out Robert's manipulation:

$$\eqalign{ & \sqrt {x\sqrt x - x} = 1 - x \cr & x\sqrt x - x = {\left( {1 - x} \right)^2} \cr & x\left( {\sqrt x - 1} \right) = {\left( {1 - x} \right)^2} \cr & \sqrt x - 1 = \frac{{{{\left( {1 - x} \right)}^2}}}{x} \cr & \sqrt x = \frac{{{{\left( {1 - x} \right)}^2}}}{x} + 1 \cr & x = {\left( {\frac{{{{\left( {1 - x} \right)}^2}}}{x} + 1} \right)^2} \cr & x = {\left( {\frac{{1 - 2x + {x^2}}}{x} + 1} \right)^2} \cr & x = {\left( {\frac{1}{x} + x - 1} \right)^2} \cr & x = {x^2} - 2x + 3 - \frac{2}{x} + \frac{1}{{{x^2}}} \cr & {x^3} = {x^4} - 2{x^3} + 3{x^2} - 2x + 1 \cr & 0 = {x^4} - 3{x^3} + 3{x^2} - 2x + 1 \cr & 0 = \left( {x - 1} \right)\left( {{x^3} - 2{x^2} + x + 1} \right) \cr} $$

Note you will most probably have two complex solutions apart from $x=1$.

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complex solutions? but the square root function is not well-defined on the complex plane! – the L May 10 '12 at 8:00
It isn't well defined in $\Bbb R$ either. We choose the positive branch. I think we can manage to do so in $\Bbb C$ too. – Pedro Tamaroff May 10 '12 at 14:45

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