Different algebraic expressions for the same roots of a polynomial.

Question

The polynomial $X^4-10X^2+1$ has roots $\pm\sqrt 2\pm\sqrt 3$. My CAS, however, returns the roots as $\pm\sqrt{5\pm2\sqrt 6}$. Probably because it uses the quadratic formula (after substituting $X^2$). Apparently, there are different algebraic expressions for the same roots and I wonder how I can obtain these on my own. In hindsight, I can convince myself that they are indeed equal, e.g. by squaring both sides of the equation $\sqrt{5+2\sqrt 6}=\sqrt 2+\sqrt 3$.

But is there a way to algebraically manipulate $\sqrt{5+2\sqrt 6}$ to directly transform it into $\sqrt2+\sqrt3$?

Yes. A square root can only be removed if the inner function is a square. So express $5 + 2\sqrt{6}$ as a square. One way is to let $\sqrt{5 + 2\sqrt{6}} = \sqrt{a} + \sqrt{b}, \space a, b \in \mathbb{N}$. Then, $$5 + 2\sqrt{6} = (\sqrt{a} + \sqrt{b})^2 = a + 2\sqrt{ab} + b$$
Now compare like terms and solve for $a, b$. — Dstarred, Jun 05 '23 at 15:21
The general tactic is to form a square. In this case you have: $\sqrt{5+2\sqrt{6}}=\sqrt{3+2\sqrt{3}\sqrt{2}+2}=\sqrt{(\sqrt{3}+\sqrt{2})^2}=\sqrt{3}+\sqrt{2}$ — Fotis, Jun 05 '23 at 15:21
Does this answer your question? Convert from Nested Square Roots to Sum of Square Roots (it contains moreover links to other duplicates) — Anne Bauval, Jun 05 '23 at 15:27
Thank you all for the quick answers! I can work with that :) If my post is a duplicate, feel free to close it. — YellowCake, Jun 05 '23 at 15:32

score 1 · Accepted Answer · answered Jun 05 '23 at 15:19

1

$$\sqrt{5+2\sqrt6}=\sqrt{(\sqrt2)^2+(\sqrt3)^2+2\sqrt2\cdot\sqrt3}=\sqrt{(\sqrt2+\sqrt3)^2}=\sqrt2+\sqrt3$$

answered Jun 05 '23 at 15:19

MathFail

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Different algebraic expressions for the same roots of a polynomial.

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