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Question Number 8351 by Nayon last updated on 09/Oct/16

     x^2 =816−64(√(77))     x=?[write x like a(√(c ))+b(√d])

$$ \\ $$$$\:\:\:{x}^{\mathrm{2}} =\mathrm{816}−\mathrm{64}\sqrt{\mathrm{77}} \\ $$$$\:\:\:{x}=?\left[{write}\:{x}\:{like}\:{a}\sqrt{{c}\:}+{b}\sqrt{\left.{d}\right]}\right. \\ $$$$ \\ $$

Commented by ridwan balatif last updated on 09/Oct/16

x=±(√(816−2×32(√(77))))  x=±(√(816−2(√(32^2 ×77))))  x=±(√(816−2(√(78848))))  x=±(√((704+112)−2(√(704×112))))  x=±((√(704))−(√(112)))  x=±(8(√(11))−4(√7))  x=±4(2(√(11))−(√7))

$$\mathrm{x}=\pm\sqrt{\mathrm{816}−\mathrm{2}×\mathrm{32}\sqrt{\mathrm{77}}} \\ $$$$\mathrm{x}=\pm\sqrt{\mathrm{816}−\mathrm{2}\sqrt{\mathrm{32}^{\mathrm{2}} ×\mathrm{77}}} \\ $$$$\mathrm{x}=\pm\sqrt{\mathrm{816}−\mathrm{2}\sqrt{\mathrm{78848}}} \\ $$$$\mathrm{x}=\pm\sqrt{\left(\mathrm{704}+\mathrm{112}\right)−\mathrm{2}\sqrt{\mathrm{704}×\mathrm{112}}} \\ $$$$\mathrm{x}=\pm\left(\sqrt{\mathrm{704}}−\sqrt{\mathrm{112}}\right) \\ $$$$\mathrm{x}=\pm\left(\mathrm{8}\sqrt{\mathrm{11}}−\mathrm{4}\sqrt{\mathrm{7}}\right) \\ $$$$\mathrm{x}=\pm\mathrm{4}\left(\mathrm{2}\sqrt{\mathrm{11}}−\sqrt{\mathrm{7}}\right) \\ $$

Answered by Rasheed Soomro last updated on 09/Oct/16

Let(a(√(c ))+b(√d))^2 = 816−64(√(77))      a,b,c,d∈N       (a(√c))^2 +2(a(√c))(b(√d))+(b(√d))^2 =816−64(√(77))       a^2 c+2ab(√(cd))+b^2 d=816−64(√(77))  Comparing          a^2 c+b^2 d=816  ∧  2ab=64  ∧ cd=77  cd=77=7×11⇒  Let c=7 , d=11   a^2 c+b^2 d=816⇒7a^2 +11b^2 =816  2ab=64⇒b=32/a  7a^2 +11b^2 =816⇒7a^2 +11(32/a)^2 =816  7a^2 +11(1024/a^2 )=816  7a^2 +11264/a^2 =816  7a^4 −816a^2 +11264=0  a^2 =((−(−816)±(√((−816)^2 −4(7)(11264))))/(2(7)))  Taking + out of ± we will get a fractional  number, which is against our assumption.  Hence we will take − sign out of ±.  a^2 =((816−(√((−816)^2 −4(7)(11264))))/(14))=((816−592)/(14))=16  a=±4⇒b=32/a=32/±4=±8  a(√(c ))+b(√d)=(±4)(√7)±8(√(11))=±4((√7)+2(√(11)))

$$\mathrm{Let}\left({a}\sqrt{{c}\:}+{b}\sqrt{{d}}\right)^{\mathrm{2}} =\:\mathrm{816}−\mathrm{64}\sqrt{\mathrm{77}}\:\:\:\:\:\:{a},{b},{c},{d}\in\mathbb{N} \\ $$$$\:\:\:\:\:\left({a}\sqrt{{c}}\right)^{\mathrm{2}} +\mathrm{2}\left({a}\sqrt{{c}}\right)\left({b}\sqrt{{d}}\right)+\left({b}\sqrt{{d}}\right)^{\mathrm{2}} =\mathrm{816}−\mathrm{64}\sqrt{\mathrm{77}} \\ $$$$\:\:\:\:\:{a}^{\mathrm{2}} {c}+\mathrm{2}{ab}\sqrt{{cd}}+{b}^{\mathrm{2}} {d}=\mathrm{816}−\mathrm{64}\sqrt{\mathrm{77}} \\ $$$${Comparing} \\ $$$$\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {d}=\mathrm{816}\:\:\wedge\:\:\mathrm{2ab}=\mathrm{64}\:\:\wedge\:\mathrm{cd}=\mathrm{77} \\ $$$$\mathrm{cd}=\mathrm{77}=\mathrm{7}×\mathrm{11}\Rightarrow\:\:{L}\mathrm{et}\:{c}=\mathrm{7}\:,\:{d}=\mathrm{11} \\ $$$$\:{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {d}=\mathrm{816}\Rightarrow\mathrm{7}{a}^{\mathrm{2}} +\mathrm{11}{b}^{\mathrm{2}} =\mathrm{816} \\ $$$$\mathrm{2}{ab}=\mathrm{64}\Rightarrow{b}=\mathrm{32}/{a} \\ $$$$\mathrm{7}{a}^{\mathrm{2}} +\mathrm{11}{b}^{\mathrm{2}} =\mathrm{816}\Rightarrow\mathrm{7}{a}^{\mathrm{2}} +\mathrm{11}\left(\mathrm{32}/{a}\right)^{\mathrm{2}} =\mathrm{816} \\ $$$$\mathrm{7}{a}^{\mathrm{2}} +\mathrm{11}\left(\mathrm{1024}/{a}^{\mathrm{2}} \right)=\mathrm{816} \\ $$$$\mathrm{7}{a}^{\mathrm{2}} +\mathrm{11264}/{a}^{\mathrm{2}} =\mathrm{816} \\ $$$$\mathrm{7}{a}^{\mathrm{4}} −\mathrm{816}{a}^{\mathrm{2}} +\mathrm{11264}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} =\frac{−\left(−\mathrm{816}\right)\pm\sqrt{\left(−\mathrm{816}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{7}\right)\left(\mathrm{11264}\right)}}{\mathrm{2}\left(\mathrm{7}\right)} \\ $$$${Taking}\:+\:{out}\:{of}\:\pm\:\mathrm{we}\:{will}\:{get}\:{a}\:{fractional} \\ $$$${number},\:{which}\:{is}\:{against}\:{our}\:{assumption}. \\ $$$${Hence}\:{we}\:{will}\:{take}\:−\:{sign}\:{out}\:{of}\:\pm. \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{816}−\sqrt{\left(−\mathrm{816}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{7}\right)\left(\mathrm{11264}\right)}}{\mathrm{14}}=\frac{\mathrm{816}−\mathrm{592}}{\mathrm{14}}=\mathrm{16} \\ $$$${a}=\pm\mathrm{4}\Rightarrow{b}=\mathrm{32}/{a}=\mathrm{32}/\pm\mathrm{4}=\pm\mathrm{8} \\ $$$${a}\sqrt{{c}\:}+{b}\sqrt{{d}}=\left(\pm\mathrm{4}\right)\sqrt{\mathrm{7}}\pm\mathrm{8}\sqrt{\mathrm{11}}=\pm\mathrm{4}\left(\sqrt{\mathrm{7}}+\mathrm{2}\sqrt{\mathrm{11}}\right) \\ $$

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