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Question Number 7611 by Tawakalitu. last updated on 06/Sep/16

Find the square root of   121x^6  + 44x^5  − 18x^4  + 18x^3  + 5x^2  − 2x + 1

$${Find}\:{the}\:{square}\:{root}\:{of}\: \\ $$$$\mathrm{121}{x}^{\mathrm{6}} \:+\:\mathrm{44}{x}^{\mathrm{5}} \:−\:\mathrm{18}{x}^{\mathrm{4}} \:+\:\mathrm{18}{x}^{\mathrm{3}} \:+\:\mathrm{5}{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{1} \\ $$

Answered by Rasheed Soomro last updated on 07/Sep/16

By division method  (√(121x^6  + 44x^5  − 18x^4  + 18x^3  + 5x^2  − 2x + 1))      { (,(11x^3 +2x^2 −x+1)),((11x^3 ),(    121x^6  + 44x^5  − 18x^4  + 18x^3  + 5x^2  − 2x + 1)),((11x^3 ),( _− 121x^6 )),((22x^3 +2x^2 ),(    44x^5 −18x^4  + 18x^3  + 5x^2  − 2x + 1)),((          +2x^2 ),(  _− 44x^5 +_(−)  4x^4  )),((22x^3 +4x^2 −x),( −22x^4 +18x^3 +5x^2 −2x+1)),((                    −x),( −_(+) 22x^4 −_(+) 4x^3  +_(−)  x^2 )),((22x^3 +4x^2 −2x+1),(    22x^3 +4x^2 −2x+1)),((                              +1),(  _− 22x^3 +_(−) 4x^2 −_(+) 2x+_(−) 1)),((22x^3 +4x^2 −2x+2),(   0     +    0    +  0 + 0    )) :}

$${By}\:{division}\:{method} \\ $$$$\sqrt{\mathrm{121}{x}^{\mathrm{6}} \:+\:\mathrm{44}{x}^{\mathrm{5}} \:−\:\mathrm{18}{x}^{\mathrm{4}} \:+\:\mathrm{18}{x}^{\mathrm{3}} \:+\:\mathrm{5}{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{1}}\: \\ $$$$ \\ $$$$\begin{cases}{}&{\mathrm{11}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{1}}\\{\mathrm{11}{x}^{\mathrm{3}} }&{\:\:\:\:\mathrm{121}{x}^{\mathrm{6}} \:+\:\mathrm{44}{x}^{\mathrm{5}} \:−\:\mathrm{18}{x}^{\mathrm{4}} \:+\:\mathrm{18}{x}^{\mathrm{3}} \:+\:\mathrm{5}{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{1}}\\{\mathrm{11}{x}^{\mathrm{3}} }&{\:_{−} \mathrm{121}{x}^{\mathrm{6}} }\\{\mathrm{22}{x}^{\mathrm{3}} +\mathrm{2}{x}^{\mathrm{2}} }&{\:\:\:\:\mathrm{44}{x}^{\mathrm{5}} −\mathrm{18}{x}^{\mathrm{4}} \:+\:\mathrm{18}{x}^{\mathrm{3}} \:+\:\mathrm{5}{x}^{\mathrm{2}} \:−\:\mathrm{2}{x}\:+\:\mathrm{1}}\\{\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{x}^{\mathrm{2}} }&{\:\:_{−} \mathrm{44}{x}^{\mathrm{5}} \underset{−} {+}\:\mathrm{4}{x}^{\mathrm{4}} \:}\\{\mathrm{22}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −{x}}&{\:−\mathrm{22}{x}^{\mathrm{4}} +\mathrm{18}{x}^{\mathrm{3}} +\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{x}}&{\:\underset{+} {−}\mathrm{22}{x}^{\mathrm{4}} \underset{+} {−}\mathrm{4}{x}^{\mathrm{3}} \:\underset{−} {+}\:{x}^{\mathrm{2}} }\\{\mathrm{22}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}&{\:\:\:\:\mathrm{22}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{1}}&{\:\:_{−} \mathrm{22}{x}^{\mathrm{3}} \underset{−} {+}\mathrm{4}{x}^{\mathrm{2}} \underset{+} {−}\mathrm{2}{x}\underset{−} {+}\mathrm{1}}\\{\mathrm{22}{x}^{\mathrm{3}} +\mathrm{4}{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}}&{\:\:\:\mathrm{0}\:\:\:\:\:+\:\:\:\:\mathrm{0}\:\:\:\:+\:\:\mathrm{0}\:+\:\mathrm{0}\:\:\:\:}\end{cases} \\ $$$$ \\ $$

Commented by Tawakalitu. last updated on 06/Sep/16

Am with you sir. thanks for your help

$${Am}\:{with}\:{you}\:{sir}.\:{thanks}\:{for}\:{your}\:{help} \\ $$

Commented by Tawakalitu. last updated on 07/Sep/16

I really appreciate . thanks so much.

$${I}\:{really}\:{appreciate}\:.\:{thanks}\:{so}\:{much}. \\ $$

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