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3 votes
Which of the following is the solution set of the equation

10(x^2+1/x^20)-63(x-1/x)=52=0
a) (5,-2,-1/5,1/2)
b) (-5,2,1/5,-1/2)
c) (5,2,-1/5,-1/2)
d) (-5,-2,-1/5,-1/2)

User TheVinchi
by
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1 Answer

2 votes
Which of the following is the solution set of the equation
10(x^2+1/x^20)-63(x-1/x)=52=0
I think you type wrongly the equation:
It is : 10(x^2+1/x^2)-63(x-1/x)+52=0
We know: x
²+1/x²=(x-1/x)²+2, pose x-1/x=t and we have x²+1/x²=t²+2
So 10(t²+2)-63t+52=0
or 10t²-63t+72=0
we have t=(63+33):20= 24/5
and t=(63-33):20=3/2

+ If t=3/2 we have x-1/x=3/2 or 2x²-3x-2=0 and we have x=2, x=-1/2
+ Of t=24/5 we have x-1/x=24/5 or 5x²-24x-5=0 or x=5 and x=-1/5

And the answer: c) (5,2,-1/5,-1/2)
User Saikamesh
by
6.3k points
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