How can I slove: $12-x=\sqrt{12-\sqrt{x}}$?
I tried to put $t=12-\sqrt{x}$
But it got me to polynomial of 4th degree which I don't think its the idea of solving this equation.
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first to the power of 2 ;then solve for x ; then check for right answer . you have to solve 4 degree equation . can you solve it by graphing ? x=9 – Khosrotash Feb 05 '15 at 20:32
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I can see that $x=9$ is a solution but how do you know that its the only one? And how can I graph it? – falcon Feb 05 '15 at 20:39
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$$f(x)=12-x\g(x)=\sqrt{12-\sqrt{x}}$$graph them , intersection(s) point are answer – Khosrotash Feb 05 '15 at 20:41
2 Answers
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$$x=12-\sqrt{12-\sqrt{x}}$$ $$x=12-\sqrt{12-\sqrt{12-\sqrt{12-\sqrt{12-\sqrt{...}}}}}$$ the value of $$\sqrt{12-\sqrt{12-\sqrt{12-\sqrt{......}}}}=3$$ $x=12-3=9$
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If $y = 12 - \sqrt{x}$, the equation says $x = 12 - \sqrt{y}$. Note the symmetry. One possibility is $x=y$ which leads to the equation $x = (12-x)^2$. That quadratic has two solutions $x=9$ and $x=16$, but only $x=9$ satisfies the original equation.
Robert Israel
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The other possibility is $y \ne x$, which leads to a quartic $x^4-48 x^3+840 x^2-6337 x+17424 = 0$, but it's divisible by the quadratic $(x-9)(x-16)$, leaving a quadratic quotient $x^2-23 x+121 $. But the roots of this quadratic don't satisfy the original equation: again they're using a wrong square root. – Robert Israel Feb 06 '15 at 00:03
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