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Question

The real numbers x1,x2,x3 satisfying the equation
x3+x2+βx+γ=0 are in A.P. Find the intervals in which β are γ lie.

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Solution

Since x1,x2,x3 are in A.P.
Let x1=ad,x2=a and x3=a+d and x1,x2,x3 are the roots of
x3x2+βx+γ=0
α=ad+a+a+d=1
a=13 .......(1)
αβ=(ad)a+a(a+d)+(ad)(a+d)=β ........(2)
and αβγ=(ad)a(a+d)=γ ........(3)
From eqn(1) we get
3a2d2=β
3(13)2d2=β
13β=d2
Since d20 for all dR
13β0
β13
β(,13]
From eqn(3),a(a2d2)=γ
13(19d2)=γ
12713d2=γ
γ+127=13d2
γ+1270
γ127
γ[127,)
Hence,β(,13] and γ[127,)

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