Elitmus
Exam
Numerical Ability
Complex Numbers
a4+1/a4=119, then a3-1/a3=????
Read Solution (Total 6)
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- ANSWER==>36
a^4+1/a^4 = (a^2+1/a^2)^2 - 2*a^2*1/a^2 =119
or (a^2+1/a^2)^2 -2 = 119
or (a^2+1/a^2)^2 = 121
hence a^2+1/a^2 = 11
again (a-1/a)^2 +2*a*1/a = 11
hence (a-1/a) = 3
now a^3-1/a^3 = (a-1/a)(a^2+1/a^2+a*1/a) = 3*(11+1) = 3*12 = 36 - 10 years agoHelpfull: Yes(35) No(0)
- (a^2+1/a^2)^2=(a^4+1/a^4+2)
hence a^2+1/a^2=11
(a-1/a)^2=a^2+1/a^2-2
a-1/a=3
(a-1/a)^3=a^3-1/a^3-3(a-1/a)
a3+1/a3=27+9=36 - 10 years agoHelpfull: Yes(4) No(0)
- @pronoy bej what about the(a-1/a)=-3???how can u ignore this value??
- 10 years agoHelpfull: Yes(1) No(1)
- a3-1/a3=(a-1/a)(a2 + 1+ 1/a2)........(1
simplify the given equation
(a2)^2 + (1/a2)^2+2*a2*1/a2=121
(a2+1/a2)=11...........(2) hence a2+1/a2+1=11+1=12;
a2+1/a2- 2*a*1/a=11-2
(a-1/a)^2=9
a-1/a=3......
use these values in eq1
3*12=36 - 10 years agoHelpfull: Yes(1) No(0)
- a^4+1/a^4=(a^2+1/a^2)^2-2.a.1/a=119
=>a^2+1/a^2=+-11
Both are square terms so we take only +11
Now,(a^2+1/a^2)=11
(a-1/a)^2+2.a.1/a=11
(a-1/a)=+-3
Now (a^3-1/a^3)
=(a-1/a)(a^2+1+1/a^2)
=+or-3 multiplied by 12
= +/-36 - 10 years agoHelpfull: Yes(0) No(2)
- a4+1/a4=(a2 + 1/a2)^2 -2=119
(a2 + 1/a2)^2=121=11
(a - 1/a)^2-2=11
(a - 1/a)=3;
now a3-1/a3=(a-1/a)^3+3(a-1/a)=3^3+3*3=36
- 9 years agoHelpfull: Yes(0) No(0)
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