Question Number 90772 by MWSuSon last updated on 26/Apr/20
$${show}\:{that}\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}=\left({b}−{c}\right)^{\mathrm{2}} −\mathrm{1}\:{are}\:{rational}\:{if} \\ $$$${b}\:{and}\:{c}\:{are}\:{rational}\:{numbers}. \\ $$
Commented by jagoll last updated on 26/Apr/20
$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\:=\:\left({b}−{c}\right)^{\mathrm{2}} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\left({x}−\mathrm{1}+{b}−{c}\right)\left({x}−\mathrm{1}+{c}−{b}\right)\:=\:\mathrm{0} \\ $$$${x}_{\mathrm{1}} \:=\:{c}+\mathrm{1}−{b}\:\:\&\:{x}_{\mathrm{2}} \:=\:{b}+\mathrm{1}−{c}\: \\ $$$${if}\:{b},{c}\:{rational}\:{number}\:{then}\: \\ $$$${x}_{\mathrm{1}} \:\wedge\:{x}_{\mathrm{2}} \:{rational}\:{number} \\ $$
Commented by MWSuSon last updated on 26/Apr/20
$${thank}\:{you}\:{sir},\:{i}\:{appreciate} \\ $$