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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-

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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 .
Ifa2−a+2=0andx2=a6+2a4+a2thenfindx?IIT−JEEbasedquestion.Findsoln.
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
Ifa2−a+2=0andx2=a6+2a4+a2thenfindx?a2−a+2=0⇒a2=a−2⇒a3=a2−2a=(a−2)−2a=−a−2⇒a4=−a2−2a=−(a−2)−2a=−3a+2⇒a5=−3a2+2a=−3(a−2)+2a=−a+6⇒a6=−a2+6a=−(a−2)+6a=5a+2x2=a6+2a4+a2=(5a+2)+2(−3a+2)+(a−2)=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)
Ifa2−a+2=0andx2=a6+2a4+a2thenfindx?a2−a+2=0⇒a2=a−2⇒a3=a2−2a=(a−2)−2a=−a−2⇒a4=−a2−2a=−(a−2)−2a=−3a+2⇒a5=−3a2+2a=−3(a−2)+2a=−a+6⇒a6=−a2+6a=−(a−2)+6a=5a+2x2=a6+2a4+a2=(5a+2)+2(−3a+2)+(a−2)=4x=±2BY:siythokozangema…(NSQ)
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
Ifa2−a+2=0andx2=a6+2a4+a2thenfindx?a2−a+2=0⇒a2=a−2x2=a2(a4+2a2+1)=a2(a2+1)2x=±a(a2+1)=±a(a−2+1)=±a(a−1)=±(a2−a)=±(a−2−a)=∓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
Ifa2−a+2=0andx2=a6+2a4+a2thenfindx?a2−a+2=0⇒{2=a−a2a2=a−2x2=a6+2a4+a2=a6+(a−a2)a4+a2=a6+a5−a6+a2=a2(a3+1)=a2(a(a−2)+1)=a2(a2−2a+1)x=±a(a−1)=±(a2−a)=±(a−2−a)=∓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=a−2⇒a4=a2−4a+4=a2−a+2−3a+2=2−3a⇒x2=a6+4−6a+a−2=a6−5a+2a3=a2−2a=a2−a+2−a−2=−2−a⇒a6=a2+4a+4=a2−a+2+5a+2=5a+2⇒x2=5a+2−5a+2=4⇒x=+−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)2→x=a(a2+1)=a2−a+2a(a−1)=a=1±i72→a(a−1)=x(1+i72)(i7−12)=−7−14=−2=x(1−i72)(−i7−12)=84=2=x

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