Question Number 147431 by vvvv last updated on 20/Jul/21
Answered by SEKRET last updated on 21/Jul/21
$$\boldsymbol{\mathrm{a}}=\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{c}}=\boldsymbol{\mathrm{d}}=\boldsymbol{\mathrm{x}} \\ $$$$ \\ $$$$\:\:\boldsymbol{\mathrm{x}}_{\mathrm{1}} \centerdot\boldsymbol{\mathrm{x}}_{\mathrm{2}} \centerdot\boldsymbol{\mathrm{x}}_{\mathrm{3}} \centerdot\boldsymbol{\mathrm{x}}_{\mathrm{4}} =\:\boldsymbol{\mathrm{w}}=\:\boldsymbol{\mathrm{abcd}} \\ $$$$\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\boldsymbol{\mathrm{mx}}^{\mathrm{3}} +\boldsymbol{\mathrm{nx}}^{\mathrm{2}} +\boldsymbol{\mathrm{kx}}+\boldsymbol{\mathrm{w}}=\mathrm{0} \\ $$$$\:\:\:\boldsymbol{\mathrm{x}}=\:\sqrt{\mathrm{4}−\sqrt{\mathrm{5}−\boldsymbol{\mathrm{x}}}} \\ $$$$\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\mathrm{4}\:=\:−\sqrt{\mathrm{5}−\boldsymbol{\mathrm{x}}}\:\: \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} +\mathrm{16}−\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{2}} \:=\:\mathrm{5}−\boldsymbol{\mathrm{x}} \\ $$$$\:\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} −\mathrm{8}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{x}}+\mathrm{11}=\mathrm{0} \\ $$$$\:\:\boldsymbol{\mathrm{abcd}}=\mathrm{11} \\ $$$$\:\:\:\:\boldsymbol{\mathrm{ABDULAZIZ}}\:\:\:\boldsymbol{\mathrm{ABDUVALIYEV}}\:\: \\ $$
Answered by gsk2684 last updated on 21/Jul/21
$${a},{b},{c},\:{and}\:{d}\:{are}\:{the}\:{roots}\:{of}\: \\ $$$${the}\:{equation}\:{x}=\sqrt{\mathrm{4}−\sqrt{\mathrm{5}−{x}}} \\ $$$${x}^{\mathrm{2}} =\mathrm{4}−\sqrt{\mathrm{5}−{x}} \\ $$$$\sqrt{\mathrm{5}−{x}}=\mathrm{4}−{x}^{\mathrm{2}} \\ $$$$\mathrm{5}−{x}=\left(\mathrm{4}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} \\ $$$$\mathrm{5}−{x}=\mathrm{16}−\mathrm{8}{x}^{\mathrm{2}} +{x}^{\mathrm{4}} \\ $$$${x}^{\mathrm{4}} −\mathrm{8}{x}^{\mathrm{2}} +{x}+\mathrm{11}=\mathrm{0}=\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)\left({x}−{d}\right) \\ $$$$\therefore\mathrm{11}={abcd} \\ $$