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If-a-2-a-2-0-and-x-2-a-6-2a-4-a-2-then-find-x-IIT-JEE-based-question-Find-sol-n-




Question Number 203941 by Panav last updated on 02/Feb/24
If a^(2 ) −a+2=0 and x^2 =a^6 +2a^4 +a^2  then find x?  IIT−JEE based question. Find sol^n .
\boldsymbolIf\boldsymbola2\boldsymbola+2=0\boldsymboland\boldsymbolx2=\boldsymbola6+2\boldsymbola4+\boldsymbola2\boldsymbolthen\boldsymbolfind\boldsymbolx?\boldsymbolIIT\boldsymbolJEE\boldsymbolbased\boldsymbolquestion.\boldsymbolFind\boldsymbolsol\boldsymboln.
Answered by Rasheed.Sindhi last updated on 02/Feb/24
If a^(2 ) −a+2=0 and x^2 =a^6 +2a^4 +a^2  then find x?  a^(2 ) −a+2=0⇒a^2 =a−2  ⇒a^3 =a^2 −2a=(a−2)−2a=−a−2  ⇒a^4 =−a^2 −2a=−(a−2)−2a=−3a+2  ⇒a^5 =−3a^2 +2a=−3(a−2)+2a=−a+6  ⇒a^6 =−a^2 +6a=−(a−2)+6a=5a+2    x^2 =a^6 +2a^4 +a^2          =(5a+2)+2(−3a+2)+(a−2)         =4  x=±2
\boldsymbolIf\boldsymbola2\boldsymbola+2=0\boldsymboland\boldsymbolx2=\boldsymbola6+2\boldsymbola4+\boldsymbola2\boldsymbolthen\boldsymbolfind\boldsymbolx?\boldsymbola2\boldsymbola+2=0a2=a2a3=a22a=(a2)2a=a2a4=a22a=(a2)2a=3a+2a5=3a2+2a=3(a2)+2a=a+6a6=a2+6a=(a2)+6a=5a+2\boldsymbolx2=\boldsymbola6+2\boldsymbola4+\boldsymbola2=(5a+2)+2(3a+2)+(a2)=4x=±2
Commented by siyathokoza last updated on 06/Feb/24
If a^(2 ) −a+2=0 and x^2 =a^6 +2a^4 +a^2  then find x?  a^(2 ) −a+2=0⇒a^2 =a−2  ⇒a^3 =a^2 −2a=(a−2)−2a=−a−2  ⇒a^4 =−a^2 −2a=−(a−2)−2a=−3a+2  ⇒a^5 =−3a^2 +2a=−3(a−2)+2a=−a+6  ⇒a^6 =−a^2 +6a=−(a−2)+6a=5a+2    x^2 =a^6 +2a^4 +a^2          =(5a+2)+2(−3a+2)+(a−2)         =4  x=±2   BY:siythokoza ngema...(NSQ)
\boldsymbolIf\boldsymbola2\boldsymbola+2=0\boldsymboland\boldsymbolx2=\boldsymbola6+2\boldsymbola4+\boldsymbola2\boldsymbolthen\boldsymbolfind\boldsymbolx?\boldsymbola2\boldsymbola+2=0a2=a2a3=a22a=(a2)2a=a2a4=a22a=(a2)2a=3a+2a5=3a2+2a=3(a2)+2a=a+6a6=a2+6a=(a2)+6a=5a+2\boldsymbolx2=\boldsymbola6+2\boldsymbola4+\boldsymbola2=(5a+2)+2(3a+2)+(a2)=4x=±2\boldsymbolBY:\boldsymbolsiythokoza\boldsymbolngema(\boldsymbolNSQ)
Answered by Rasheed.Sindhi last updated on 02/Feb/24
If a^(2 ) −a+2=0 and x^2 =a^6 +2a^4 +a^2  then find x?  a^(2 ) −a+2=0⇒a^2 =a−2     x^2 =a^2 (a^4 +2a^2 +1)       =a^2 (a^2 +1)^2   x=±a(a^2 +1)     =±a(a−2+1)     =±a(a−1)     =±(a^2 −a)     =±(a−2−a)    =∓2  x=±2
\boldsymbolIf\boldsymbola2\boldsymbola+2=0\boldsymboland\boldsymbolx2=\boldsymbola6+2\boldsymbola4+\boldsymbola2\boldsymbolthen\boldsymbolfind\boldsymbolx?\boldsymbola2\boldsymbola+2=0a2=a2x2=a2(a4+2a2+1)=a2(a2+1)2x=±a(a2+1)=±a(a2+1)=±a(a1)=±(a2a)=±(\cancela2\cancela)=2x=±2
Commented by Panav last updated on 10/Feb/24
Correct
Correct
Answered by Rasheed.Sindhi last updated on 02/Feb/24
If a^(2 ) −a+2=0 and x^2 =a^6 +2a^4 +a^2  then find x?  a^(2 ) −a+2=0⇒ { ((2=a−a^2 )),((a^2 =a−2)) :}     x^2 =a^6 +2a^4 +a^2 =a^6 +(a−a^2 )a^4 +a^2       =a^6 +a^5 −a^6 +a^2 =a^2 (a^3 +1)     =a^2 ( a(a−2)+1 )     =a^2 (a^2 −2a+1)  x =±a(a−1)      =±(a^2 −a)      =±(a−2−a)      =∓2  x=±2
\boldsymbolIf\boldsymbola2\boldsymbola+2=0\boldsymboland\boldsymbolx2=\boldsymbola6+2\boldsymbola4+\boldsymbola2\boldsymbolthen\boldsymbolfind\boldsymbolx?\boldsymbola2\boldsymbola+2=0{2=aa2a2=a2\boldsymbolx2=\boldsymbola6+2\boldsymbola4+\boldsymbola2=a6+(aa2)a4+a2=\cancela6+a5\cancela6+a2=a2(a3+1)=a2(a(a2)+1)=a2(a22a+1)x=±a(a1)=±(a2a)=±(a2a)=2x=±2
Answered by AST last updated on 02/Feb/24
a^2 =a−2⇒a^4 =a^2 −4a+4=a^2 −a+2−3a+2=2−3a  ⇒x^2 =a^6 +4−6a+a−2=a^6 −5a+2  a^3 =a^2 −2a=a^2 −a+2−a−2=−2−a  ⇒a^6 =a^2 +4a+4=a^2 −a+2+5a+2=5a+2  ⇒x^2 =5a+2−5a+2=4⇒x=+_− 2
a2=a2a4=a24a+4=a2a+23a+2=23ax2=a6+46a+a2=a65a+2a3=a22a=a2a+2a2=2aa6=a2+4a+4=a2a+2+5a+2=5a+2x2=5a+25a+2=4x=+2
Answered by esmaeil last updated on 02/Feb/24
x^2 =(a^3 +a)^2 →x=a(a^2 +1)=^(a^2 −a+2) a(a−1)=  a=((1±i(√7))/2)→a(a−1)=x  (((1+i(√7))/2))(((i(√7)−1)/2))=((−7−1)/4)=−2=x  (((1−i(√7))/2))(((−i(√7)−1)/2))=(8/4)=2=x
x2=(a3+a)2x=a(a2+1)=a2a+2a(a1)=a=1±i72a(a1)=x(1+i72)(i712)=714=2=x(1i72)(i712)=84=2=x

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