Question Number 216207 by MATHEMATICSAM last updated on 30/Jan/25
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{any}\:\mathrm{kind}\:\mathrm{of}\:\mathrm{equation}\:\mathrm{should} \\ $$$$\mathrm{have}\:\mathrm{atleast}\:\mathrm{one}\:\mathrm{root}.\:\left(\mathrm{Algebric}\:\right. \\ $$$$\left.\mathrm{fundamental}\:\mathrm{theorem}\right) \\ $$
Commented by Ghisom last updated on 30/Jan/25
$$\mathrm{not}\:{any}\:{kind}\:{of}\:{equation}.\:\mathrm{for}\:\mathrm{example} \\ $$$$\sqrt{{x}}=−\mathrm{1}\:\mathrm{has}\:\mathrm{no}\:\mathrm{solution}\:\mathrm{at}\:\mathrm{all}. \\ $$$$ \\ $$$$\mathrm{the}\:\mathrm{theorem}\:\mathrm{states}\:\mathrm{that}: \\ $$$$ \\ $$$$“\mathrm{every}\:\mathrm{non}−\mathrm{constant}\:\mathrm{single}−\mathrm{variable} \\ $$$$\mathrm{polynomial}\:\mathrm{with}\:\mathrm{complex}\:\mathrm{coefficients}\:\mathrm{has} \\ $$$$\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{complex}\:\mathrm{solution}'' \\ $$
Commented by MATHEMATICSAM last updated on 30/Jan/25
$$\mathrm{ya}\:\mathrm{maybe}\:\mathrm{but}\:\mathrm{I}\:\mathrm{want}\:\mathrm{proof} \\ $$