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Question Number 215625 by MathematicalUser2357 last updated on 12/Jan/25

x^4 +x^3 −8x^2 +2x+4=0 x=1 ∨ x=2 ∨ x=2±(√2) Is this right? I have not enough time to edit my solution

$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}=\mathrm{0} \\ $$$${x}=\mathrm{1}\:\vee\:{x}=\mathrm{2}\:\vee\:{x}=\mathrm{2}\pm\sqrt{\mathrm{2}} \\ $$$$\mathrm{Is}\:\mathrm{this}\:\mathrm{right}?\:\mathrm{I}\:\mathrm{have}\:\mathrm{not}\:\mathrm{enough}\:\mathrm{time}\:\mathrm{to}\:\mathrm{edit}\:\mathrm{my}\:\mathrm{solution} \\ $$

Commented by MathematicalUser2357 last updated on 12/Jan/25

why did you post late???

$$\mathrm{why}\:\mathrm{did}\:\mathrm{you}\:\mathrm{post}\:\mathrm{late}??? \\ $$

Commented by mr W last updated on 12/Jan/25

if you are not sleeping, you should be able to know by youself that your result is wrong, since 1+2+(2+(√2))+(2−(√2))≠−1 the correct answer should be x=1 ∨ x=2 ∨ x=−2±(√2) therefore the someone who is sleeping is actually youself.

$${if}\:{you}\:{are}\:{not}\:{sleeping},\:{you}\:{should} \\ $$$${be}\:{able}\:{to}\:{know}\:{by}\:{youself}\:{that}\: \\ $$$${your}\:{result}\:{is}\:{wrong},\:{since}\: \\ $$$$\mathrm{1}+\mathrm{2}+\left(\mathrm{2}+\sqrt{\mathrm{2}}\right)+\left(\mathrm{2}−\sqrt{\mathrm{2}}\right)\neq−\mathrm{1} \\ $$$${the}\:{correct}\:{answer}\:{should}\:{be} \\ $$$${x}=\mathrm{1}\:\vee\:{x}=\mathrm{2}\:\vee\:{x}=−\mathrm{2}\pm\sqrt{\mathrm{2}} \\ $$$${therefore}\:{the}\:{someone}\:{who}\:{is}\: \\ $$$${sleeping}\:{is}\:{actually}\:{youself}. \\ $$

Commented by MathematicalUser2357 last updated on 12/Jan/25

I think someone is sleeping instead of answering

$$\mathrm{I}\:\mathrm{think}\:\mathrm{someone}\:\mathrm{is}\:\mathrm{sleeping}\:\mathrm{instead}\:\mathrm{of}\:\mathrm{answering} \\ $$

Commented by mr W last updated on 12/Jan/25

nobody must answer you at all! if you seriously need help, just post your questions and wait. weird comments like those above are just annoying. they help neither youself nor others.

$${nobody}\:{must}\:{answer}\:{you}\:{at}\:{all}! \\ $$$${if}\:{you}\:{seriously}\:{need}\:{help},\:{just}\:{post} \\ $$$${your}\:{questions}\:{and}\:{wait}.\:{weird} \\ $$$${comments}\:{like}\:{those}\:{above}\:{are}\:{just}\: \\ $$$${annoying}.\:{they}\:{help}\:{neither}\:{youself} \\ $$$${nor}\:{others}. \\ $$

Commented by MathematicalUser2357 last updated on 13/Jan/25

I failed in x=2±(√2)

$$\mathrm{I}\:\mathrm{failed}\:\mathrm{in}\:{x}=\mathrm{2}\pm\sqrt{\mathrm{2}} \\ $$

Answered by MathematicalUser2357 last updated on 13/Jan/25

x^4 +x^3 −8x^2 +2x+4=0 P(x)=x^4 +x^3 −8x^2 +2x+4⇒P(1)=0 determinant (((Order),(Dividend/Remainder),(Divisor),(Quotient),(Removal)),(1,(x^4 +x^3 −8x^2 +2x+4),(x−1),x^3 ,(x^4 −x^3 )),(2,(2x^3 −8x^2 +2x+4),,(2x^2 ),(2x^3 −2x^2 )),(3,(−6x^2 +2x+4),,(−6x),(−6x^2 +6x)),(4,(−4x+4),,(−4),(−4x+4)),((Total),0,,(x^3 +2x^2 −6x−4),(x^4 +x^3 −8x^2 +2x+4))) ∴P(x)=(x−1)(x^3 +2x^2 −6x−4)⇒x=1 Q(x)=x^3 +2x^2 −6x−4⇒Q(2)=0 determinant (((Order),(Dividend/Remainder),(Divisor),(Quotient),(Removal)),(1,(x^3 +2x^2 −6x−4),(x−2),x^2 ,(x^3 −2x^2 )),(2,(4x^2 −6x−4),,(4x),(4x^2 −8x)),(3,(2x−4),,2,(2x−4)),((Total),0,,(x^2 +4x+2),(x^3 +2x^2 −6x−4))) ∴Q(x)=(x−2)(x^2 +4x+2)⇒x=2 or x=−2±(√2) Welp, one mistake on mondays.

$${x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}=\mathrm{0} \\ $$$${P}\left({x}\right)={x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}\Rightarrow{P}\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\begin{array}{|c|c|c|c|c|c|}{\mathrm{Order}}&\hline{\mathrm{Dividend}/\mathrm{Remainder}}&\hline{\mathrm{Divisor}}&\hline{\mathrm{Quotient}}&\hline{\mathrm{Removal}}\\{\mathrm{1}}&\hline{{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}&\hline{{x}−\mathrm{1}}&\hline{{x}^{\mathrm{3}} }&\hline{{x}^{\mathrm{4}} −{x}^{\mathrm{3}} }\\{\mathrm{2}}&\hline{\mathrm{2}{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}&\hline{}&\hline{\mathrm{2}{x}^{\mathrm{2}} }&\hline{\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} }\\{\mathrm{3}}&\hline{−\mathrm{6}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}&\hline{}&\hline{−\mathrm{6}{x}}&\hline{−\mathrm{6}{x}^{\mathrm{2}} +\mathrm{6}{x}}\\{\mathrm{4}}&\hline{−\mathrm{4}{x}+\mathrm{4}}&\hline{}&\hline{−\mathrm{4}}&\hline{−\mathrm{4}{x}+\mathrm{4}}\\{\mathrm{Total}}&\hline{\mathrm{0}}&\hline{}&\hline{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{4}}&\hline{{x}^{\mathrm{4}} +{x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{4}}\\\hline\end{array} \\ $$$$\therefore{P}\left({x}\right)=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{4}\right)\Rightarrow{x}=\mathrm{1} \\ $$$${Q}\left({x}\right)={x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{4}\Rightarrow{Q}\left(\mathrm{2}\right)=\mathrm{0} \\ $$$$\begin{array}{|c|c|c|c|c|}{\mathrm{Order}}&\hline{\mathrm{Dividend}/\mathrm{Remainder}}&\hline{\mathrm{Divisor}}&\hline{\mathrm{Quotient}}&\hline{\mathrm{Removal}}\\{\mathrm{1}}&\hline{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{4}}&\hline{{x}−\mathrm{2}}&\hline{{x}^{\mathrm{2}} }&\hline{{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} }\\{\mathrm{2}}&\hline{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{4}}&\hline{}&\hline{\mathrm{4}{x}}&\hline{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{8}{x}}\\{\mathrm{3}}&\hline{\mathrm{2}{x}−\mathrm{4}}&\hline{}&\hline{\mathrm{2}}&\hline{\mathrm{2}{x}−\mathrm{4}}\\{\mathrm{Total}}&\hline{\mathrm{0}}&\hline{}&\hline{{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}}&\hline{{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −\mathrm{6}{x}−\mathrm{4}}\\\hline\end{array} \\ $$$$\therefore{Q}\left({x}\right)=\left({x}−\mathrm{2}\right)\left({x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{2}\right)\Rightarrow{x}=\mathrm{2}\:\mathrm{or}\:{x}=−\mathrm{2}\pm\sqrt{\mathrm{2}} \\ $$$$\mathrm{Welp},\:\mathrm{one}\:\mathrm{mistake}\:\mathrm{on}\:\mathrm{mondays}. \\ $$

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