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Algebra Level 3

x 5 − 40 x 4 + P x 3 + Q x 2 + R x + S = 0 x^5-40x^4+Px^3+Qx^2+Rx+S=0

Given that all the roots of the equation above follows a geometric progression , and the sum of reciprocals of all these roots is 10. Find ∣ S ∣ |S| .

Notation : ∣ ⋅ ∣ | \cdot | denotes the absolute value function .


The answer is 32.

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Ayush G Rai by Ayush G Rai , 20, India

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1 solution

Ayush G Rai
Jun 23, 2016

Let the roots be a r 2 , a r , a , a r , a r 2 . ar^2,ar,a,\dfrac{a}{r},\dfrac{a}{r^2}.
Then the sum of the roots = a ( r 2 + r + 1 + 1 r + 1 r 2 ) = 40 =a(r^2+r+1+\dfrac{1}{r}+\dfrac{1}{r^2})=40 according to vieta's formula.
The sum of their reciprocals = 1 a ( 1 r 2 + 1 r + 1 + r + r 2 ) = 10 =\dfrac{1}{a}(\dfrac{1}{r^2}+\dfrac{1}{r}+1+r+r^2)=10
Dividing the two equations we get, a 2 = 4 a^2=4 or a = ± 2. a=\pm2.
Product of the roots = − S = a 5 = ( ± 2 ) 5 = ± 32 =-S=a^5={(\pm 2)}^5=\pm 32
Therefore ∣ S ∣ = 32 . |S|=\boxed {32}.


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