$n^2-4n+2=0$ I have tried many things for this but I cant resolve the roots
here $n$ should be a positive whole number as it stands for time.
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3This equation has no solution over the integers. – Darth Geek Jul 26 '14 at 17:29
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1By the rational root theorem, the only possible rational roots are $\pm 1,\pm2$. None of these work. (Alternatively, note that this is the same as $(n-2)^2=2$---but you certainly can't square any integers and get 2.) – Semiclassical Jul 26 '14 at 17:31
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why was this downvoted – agha rehan abbas Jul 26 '14 at 17:53
3 Answers
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$$\Delta=(-4)^2-4 \cdot 2=16-8=8$$
$$n_{1,2}=\frac{4 \pm \sqrt{8}}{2}=\frac{4 \pm 2\sqrt{2}}{2}=2 \pm \sqrt{2}$$
So,the roots are $2 + \sqrt{2}$ and $2 - \sqrt{2}$ (both of them are positive).
evinda
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Using the discriminant $$\Delta=4^2-4\times 2=8$$ so the roots are $$n_1=\frac12(4+\sqrt8)\qquad n_2=\frac12(4-\sqrt8)$$ and these two roots are irrational.
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Hint $\ n^2- 4n+2 = 0 \,\Rightarrow\, (n-2)^2 = 2\,\Rightarrow\,n\,$ irrational
Bill Dubuque
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