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Question Number 90772 by MWSuSon last updated on 26/Apr/20

show that the roots of the equation  x^2 −2x=(b−c)^2 −1 are rational if  b and c are rational numbers.

$${show}\:{that}\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{x}=\left({b}−{c}\right)^{\mathrm{2}} −\mathrm{1}\:{are}\:{rational}\:{if} \\ $$$${b}\:{and}\:{c}\:{are}\:{rational}\:{numbers}. \\ $$

Commented by jagoll last updated on 26/Apr/20

x^2 −2x+1 = (b−c)^2   (x−1)^2 −(b−c)^2  = 0  (x−1+b−c)(x−1+c−b) = 0  x_1  = c+1−b  & x_2  = b+1−c   if b,c rational number then   x_1  ∧ x_2  rational number

$${x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}\:=\:\left({b}−{c}\right)^{\mathrm{2}} \\ $$$$\left({x}−\mathrm{1}\right)^{\mathrm{2}} −\left({b}−{c}\right)^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\left({x}−\mathrm{1}+{b}−{c}\right)\left({x}−\mathrm{1}+{c}−{b}\right)\:=\:\mathrm{0} \\ $$$${x}_{\mathrm{1}} \:=\:{c}+\mathrm{1}−{b}\:\:\&\:{x}_{\mathrm{2}} \:=\:{b}+\mathrm{1}−{c}\: \\ $$$${if}\:{b},{c}\:{rational}\:{number}\:{then}\: \\ $$$${x}_{\mathrm{1}} \:\wedge\:{x}_{\mathrm{2}} \:{rational}\:{number} \\ $$

Commented by MWSuSon last updated on 26/Apr/20

thank you sir, i appreciate

$${thank}\:{you}\:{sir},\:{i}\:{appreciate} \\ $$

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