Question-59225

Question Number 59225 by azizullah last updated on 06/May/19

Commented by malwaan last updated on 06/May/19

$${Whats}\:{the}\:{first}\:{term}\:? \\ $$$${x}^{\mathrm{4}} \:{or}\:\:{x}^{\mathrm{2}} \:? \\ $$

Commented by azizullah last updated on 06/May/19

dear sir there is mistake it taken from book but i think the term is x^4

$$\boldsymbol{\mathrm{dear}}\:\boldsymbol{\mathrm{sir}}\:\boldsymbol{\mathrm{there}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{mistake}}\:\boldsymbol{\mathrm{it}}\:\boldsymbol{\mathrm{taken}}\:\boldsymbol{\mathrm{from}}\:\boldsymbol{\mathrm{book}}\:\boldsymbol{\mathrm{but}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{think}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{term}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} \\ $$

Commented by MJS last updated on 06/May/19

$$\mathrm{no}\:\mathrm{easy}\:\mathrm{solution}\:\mathrm{in}\:\mathrm{both}\:\mathrm{cases} \\ $$

Answered by malwaan last updated on 06/May/19

(x^4 −x^3 −2x^2 +x+1)/(x−1) =x^3 −2x−1 (x^3 −2x−1)/(x+1)= x^2 −x−1=0 ⇒x=((1±(√5))/2) The solution set= {1 ; −1 ; ((1+(√5))/2) ; ((1−(√5))/2) }

$$\left({x}^{\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)/\left({x}−\mathrm{1}\right) \\ $$$$\:={x}^{\mathrm{3}} −\mathrm{2}{x}−\mathrm{1} \\ $$$$\left({x}^{\mathrm{3}} −\mathrm{2}{x}−\mathrm{1}\right)/\left({x}+\mathrm{1}\right)= \\ $$$${x}^{\mathrm{2}} −{x}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$${The}\:{solution}\:{set}= \\ $$$$\left\{\mathrm{1}\:;\:−\mathrm{1}\:;\:\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:;\:\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\:\right\} \\ $$

Commented by MJS last updated on 06/May/19

$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:{x}^{\mathrm{2}\vee\mathrm{4}} −{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$

Commented by malwaan last updated on 07/May/19

$${yes}\:{you}\:{are}\:{right} \\ $$$${I}\:{am}\:{sorry} \\ $$

Commented by azizullah last updated on 07/May/19

$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{Now}}\:\boldsymbol{\mathrm{solve}}\:\boldsymbol{\mathrm{x}}^{\mathrm{4}} −\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{3}} −\mathrm{2}\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\mathrm{2}\boldsymbol{\mathrm{x}}+\mathrm{1}=\:\mathrm{0} \\ $$

Commented by MJS last updated on 07/May/19

x^4 −2x^3 −2x^2 +2x+1=0 (x^4 −2x^2 +1)+(−2x^3 +2x)=0 (x^2 −1)^2 −2x(x^2 −1)=0 (x^2 −1)((x^2 −1)−2x)=0 (x−1)(x+1)(x^2 −2x−1)=0 ⇒ x_(1, 2) =±1 x_(3, 4) =1±(√2)

$${x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)+\left(−\mathrm{2}{x}^{\mathrm{3}} +\mathrm{2}{x}\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} −\mathrm{2}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left(\left({x}^{\mathrm{2}} −\mathrm{1}\right)−\mathrm{2}{x}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{1}\right)\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}_{\mathrm{1},\:\mathrm{2}} =\pm\mathrm{1} \\ $$$$\:\:\:\:\:\:{x}_{\mathrm{3},\:\mathrm{4}} =\mathrm{1}\pm\sqrt{\mathrm{2}} \\ $$

Commented by azizullah last updated on 07/May/19

$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{so}}\:\boldsymbol{\mathrm{much}}\:\boldsymbol{\mathrm{answer}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{correct}} \\ $$

Commented by malwaan last updated on 09/May/19

(x^4 −2x^3 −2x^2 +2x+1)/(x−1) =x^3 −x^2 −3x−1 (x^3 −x^2 −3x−1)/(x+1) =x^2 −2x−1 ⇒x=((+2±(√(4+4)))/2)=1±(√2) ∴x_1 =1 ; x_2 =−1 x_3 =1+(√2) ; x_4 =1−(√2)

$$\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{3}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right)/\left({x}−\mathrm{1}\right) \\ $$$$={x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1} \\ $$$$\left({x}^{\mathrm{3}} −{x}^{\mathrm{2}} −\mathrm{3}{x}−\mathrm{1}\right)/\left({x}+\mathrm{1}\right) \\ $$$$={x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1} \\ $$$$\Rightarrow{x}=\frac{+\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{4}}}{\mathrm{2}}=\mathrm{1}\pm\sqrt{\mathrm{2}} \\ $$$$\therefore{x}_{\mathrm{1}} =\mathrm{1}\:;\:\:{x}_{\mathrm{2}} =−\mathrm{1} \\ $$$${x}_{\mathrm{3}} =\mathrm{1}+\sqrt{\mathrm{2}}\:\:;\:\:{x}_{\mathrm{4}} =\mathrm{1}−\sqrt{\mathrm{2}} \\ $$

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