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If-x-1-i-is-a-root-of-the-equation-x-3-ix-1-i-0-then-the-other-real-root-is-




Question Number 49460 by Pk1167156@gmail.com last updated on 07/Dec/18
If  x=1+i  is a root of the equation  x^3 −ix+1−i=0 , then the other real  root is
$$\mathrm{If}\:\:{x}=\mathrm{1}+{i}\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$${x}^{\mathrm{3}} −{ix}+\mathrm{1}−{i}=\mathrm{0}\:,\:\mathrm{then}\:\mathrm{the}\:\mathrm{other}\:\mathrm{real} \\ $$$$\mathrm{root}\:\mathrm{is} \\ $$
Commented by Kunal12588 last updated on 07/Dec/18
i do it with normal real way. so it  should be wrong.  x^3 −ix+1−i=(x−1−i)(x+i)(x+1)  roots are  x_1 =1+i,x_2 =−i,x_3 =−1. so real root=−1.  please inform if it is wrong.
$${i}\:{do}\:{it}\:{with}\:{normal}\:{real}\:{way}.\:{so}\:{it} \\ $$$${should}\:{be}\:{wrong}. \\ $$$${x}^{\mathrm{3}} −{ix}+\mathrm{1}−{i}=\left({x}−\mathrm{1}−{i}\right)\left({x}+{i}\right)\left({x}+\mathrm{1}\right) \\ $$$${roots}\:{are} \\ $$$${x}_{\mathrm{1}} =\mathrm{1}+{i},{x}_{\mathrm{2}} =−{i},{x}_{\mathrm{3}} =−\mathrm{1}.\:{so}\:{real}\:{root}=−\mathrm{1}. \\ $$$${please}\:{inform}\:{if}\:{it}\:{is}\:{wrong}. \\ $$
Commented by Abdo msup. last updated on 08/Dec/18
(x−1−i)(x+i)(x+1)=(x−1−i)(x^2 +x+ix +i)  =(x−1−i)(x^2  +(1+i)x +i)  =x^3  +(1+i)x^2 +ix−x^2 −(1+i)x−i−ix^2 −i(1+i)x +1  x^3  −x −ix +x +1−i = x^3 −ix +1−i so your answer is  correct sir .
$$\left({x}−\mathrm{1}−{i}\right)\left({x}+{i}\right)\left({x}+\mathrm{1}\right)=\left({x}−\mathrm{1}−{i}\right)\left({x}^{\mathrm{2}} +{x}+{ix}\:+{i}\right) \\ $$$$=\left({x}−\mathrm{1}−{i}\right)\left({x}^{\mathrm{2}} \:+\left(\mathrm{1}+{i}\right){x}\:+{i}\right) \\ $$$$={x}^{\mathrm{3}} \:+\left(\mathrm{1}+{i}\right){x}^{\mathrm{2}} +{ix}−{x}^{\mathrm{2}} −\left(\mathrm{1}+{i}\right){x}−{i}−{ix}^{\mathrm{2}} −{i}\left(\mathrm{1}+{i}\right){x}\:+\mathrm{1} \\ $$$${x}^{\mathrm{3}} \:−{x}\:−{ix}\:+{x}\:+\mathrm{1}−{i}\:=\:{x}^{\mathrm{3}} −{ix}\:+\mathrm{1}−{i}\:{so}\:{your}\:{answer}\:{is} \\ $$$${correct}\:{sir}\:. \\ $$

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