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How can I remove the outer square of $ \sqrt{11 - \sqrt{3}} $, so that we don't have nested root, like we can do for $ \sqrt{4 - \sqrt{7}} $ as shown below?

$$ \sqrt{4 - \sqrt{7}} = \sqrt{(4 - \sqrt{7}) \times \frac{2}{2}} \quad \text{(multiplying by } \frac{2}{2} \text{)} \\ = \sqrt{\frac{8 - 2\sqrt{7}}{2}} \\ = \sqrt{\frac{(\sqrt{7} - \sqrt{1})^2}{2}} \quad \because 8 - 2\sqrt{7} = (\sqrt{7} - \sqrt{1})^2 \\ = \frac{\sqrt{7} - \sqrt{1}}{\sqrt{2}} \\ = \frac{\sqrt{7} - 1}{\sqrt{2}} $$

This method works for $ \sqrt{4 - \sqrt{7}} $ but not for $ \sqrt{11 - \sqrt{3}} $. I want to know a general method for it.

I need to know this because in solving inequalities like $ \sqrt{x-6} - \sqrt{10-x} \geq 1$, I get probable solutions in the form like $ \frac{ 8 \pm \sqrt{ 7 }}{ 2 } $ or $ \sqrt{4 - \sqrt{7}} $ and hence in verification of whether these probable solutions are actual solutions I get nested roots.

Thomas Andrews
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Kuldeep
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    Check https://math.stackexchange.com/questions/196155/strategies-to-denest-nested-radicals-sqrtab-sqrtc – Benjamin Wang Jun 12 '25 at 07:57
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    Is there a reason to expect you can? – Thomas Andrews Jun 12 '25 at 08:04
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    @ThomasAndrews my teacher made a passing comment while solving the mentioned inequality that we can always do such simplification, I felt he was wrong and hence came up with this problem to contradict him. – Kuldeep Jun 12 '25 at 08:12
  • Take the forth root of 7 which is the square root of the square root. If we could remove the outer we would not need forth, eights, 16nth roots at all. Not a proof, but a strong hint that your teacher is wrong, or that his comment was meant in a context of special cases. – Gyro Gearloose Jun 12 '25 at 08:34
  • It seems like, if there is a way to write it this way, it is as $a\sqrt{7n}+b\sqrt{21n},$ with $a,b$ rational and $n$ an integer . In your example, the answer can be written as $\frac12\sqrt{14}-\frac12\sqrt2.$ That's just a guess, though, because $11^2-3=118=4^2\cdot 7.$ – Thomas Andrews Jun 12 '25 at 09:02
  • But the way, your inequality example is not, in fact, an inequality. – Thomas Andrews Jun 12 '25 at 09:20
  • @ThomasAndrews oops! i fixed it. thank you. – Kuldeep Jun 12 '25 at 09:30

1 Answers1

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The general method is as follows. You're trying to write $\sqrt{a\pm \sqrt{b}}$ as $\sqrt{c}\pm \sqrt{d}$. Square both sides to get $a\pm\sqrt{b}=c+d\pm 2\sqrt{cd}$. Provided that $b$ is not a perfect square, $a$ and $\sqrt{b}$ are linearly independent over the rationals. So you need $a=c+d$ and $b=4cd$. Thus $c$ and $d$ are the solutions of $x^2-ax+\frac{1}{4}b=0$. This factorises rationally if and only if the discriminant $a^2-b$ is a perfect square. If $a=4$ and $b=7$ then $16-7$ is indeed a perfect square. If $a=11$ and $b=3$ then $121-3$ isn't.

Dave Benson
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