I am trying to simplify $c=\sqrt{290-143\sqrt2}.$ I am solving a triangle and I got that $c^2=290-143\sqrt2.$ I have tried to use the formula for $\sqrt{a\pm\sqrt{b}}$ but it seemed useless at the end. Can you give me a hint? I want to remove the square root.
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Try setting $a(290-143\sqrt 2) = (b+c\sqrt 2)^2$ and solving for $b,c$ in terms or $a$. – Vishu Jan 19 '21 at 12:04
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@KaloyanK. can you show us how you´ve obtained $c^2=290-143\sqrt2?$ Are you sure there isn't a typo? – user376343 Jan 19 '21 at 12:17
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@KaloyanK. Alright. – Vishu Jan 19 '21 at 12:19
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1I’m voting to close this question because it has been solved by comments. – Sebastiano Jan 19 '21 at 22:19
2 Answers
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It is impossible to remove the square root and get something of the form $a+b\sqrt c$ where $a,b,c$ are rational numbers. If that were possible, the minimal polynomial of the given number would be quadratic. But that of $\sqrt{290-143\sqrt2}$ is a quartic $x^4-580x^2+43202=0$.
Parcly Taxel
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Hint
$$(\sqrt{a}-\sqrt{b})^2=290-143\sqrt{2}$$ $$a+b-2\sqrt{ab}=290-143\sqrt{2}$$ so, $$a+b=290 \quad \text{and} \quad 4ab=2.143^2$$ which is impossible to solve for $a$ and $b$ rational numbers.
Arnaldo
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