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The equation is $$\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$$

I cannot find any way to simplify this, other than squaring repeatedly, which is, btw, again not simplifying. Also tried substitutions like $1-x = t^2$ and trigonometric substitutions. None seem to work.

By trial and error, I find one root to be $\dfrac{16}{25}$. Is there a general way to treat this equation?

EDIT

The domain of definition of this function seems to be quite small, some subset of $(0,1)$. Also we know that there is only one root as the function is strictly increasing.

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$$\sqrt{x-\sqrt{1-x}}=1-\sqrt{x}$$Squaring we get $$x-\sqrt{1-x}=1-2\sqrt{x}+x\\2\sqrt{x}-\sqrt{1-x}=1$$ Again squaring $$4x-4\sqrt{x(1-x)}+1-x=1\\3x=4\sqrt{x(1-x)}$$And again $$9x^2=16x(1-x)\\9x^2=16x-16x^2\\25x^2-16x=0$$ We can see that $x=0$ is not a root since $\sqrt{x-\sqrt{1-x}}$ is not defined for $x=0$ and checking we see that $x=\frac{16}{25}$ is indeed a root (as you guessed).

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  • $\begingroup$ I guess I am too sleepy to not see that. Thanks a lot. $\endgroup$ – samjoe Oct 23 '17 at 18:51
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    $\begingroup$ @samjoe No problem, happens often (that you overlook an easy solution) :) $\endgroup$ – kingW3 Oct 23 '17 at 18:56
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Repeated squaring gets me to a quadratic $$\sqrt{x-\sqrt{1-x}}=1-\sqrt x\\ x-\sqrt{1-x}=1+x-2\sqrt x\\ 2\sqrt x-\sqrt{1-x}=1\\ 4x+1-x-4\sqrt{x(1-x)}=1\\ 9x^2=16x(1-x)\\ x=\frac {16}{25},0$$ and we can check that $0$ does not work.

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  • $\begingroup$ Thanks Ross, I thought squaring would get me a more complicated equation. $\endgroup$ – samjoe Oct 23 '17 at 18:52
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Following the suggestion in comments, multiply both sides by:

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

and you get $$\sqrt{1-x}=\sqrt{x}- \sqrt{x-\sqrt{1-x}}$$

Adding to the original equation, and you get:

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

That's a little easier to solve.

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  • $\begingroup$ Thanks! This is also possible. $\endgroup$ – samjoe Oct 23 '17 at 18:54
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writing your equation in the form $$\sqrt{x-\sqrt{1-x}}=1-\sqrt{x}$$ squaring we obtain $$\sqrt{1-x}=2\sqrt{x}-1$$ squaring again $$4\sqrt{x}=5x$$ and we get $$25x^2-16x=0$$ thus $$x=0$$ or $$x=\frac{16}{25}$$

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  • $\begingroup$ But $x=0$ is not a solution... $\endgroup$ – Thomas Andrews Oct 23 '17 at 18:47
  • $\begingroup$ this is right, you must make a Control with all these "Solutions" in the given equation $\endgroup$ – Dr. Sonnhard Graubner Oct 23 '17 at 18:50
  • $\begingroup$ Thanks Sonnhard Graubner! $\endgroup$ – samjoe Oct 23 '17 at 18:53
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$$\sqrt{x} + \sqrt{x-\sqrt{1-x}} = 1$$ $$\sqrt{x-\sqrt{1-x}} = 1 -\sqrt{x}$$ $$x-\sqrt{1-x} = (1 -\sqrt{x})^2$$ $$x-\sqrt{1-x} = x- 2\sqrt{x} +1$$ $$\sqrt{1-x} = 2\sqrt{x} +1$$ $$1-x = (2\sqrt{x} +1)^2$$ $$1-x = 1 + 4 \sqrt{x} + 4 x$$ $$- 5x= 4 \sqrt{x} $$ $$- \frac{5}{4}x= \sqrt{x} $$ $$\left(- \frac{5}{4}x\right)^2= x $$ $$\frac{25}{16}x^2= x $$ $$\frac{25}{16}= \frac{1}{x} $$ $$\frac{16}{25}= x $$

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