(α,β);(β,ϒ) and (ϒ,α) are respectively the roots of x² -2px+2 =0, x² -2qx +3=0 and x² -2rx+6=0.If α,β and ϒ are all positive ,then prove that the value of p+q+r is 6

Question

(α,β);(β,ϒ) and (ϒ,α) are
respectively the roots of x2 -2px+2 =0, x2
-2qx +3=0 and x2 -2rx+6=0 If α,β and ϒ
are all positive ,then prove that the value of p+q+r is 6

Priyanka Kedia · Accepted Answer

Dear Student,
Please find below the solution to the asked
query:

Given α,β are the roots of x2-2px+2=0
 So,α+β=2p     
1αβ=2           
2And β,γ are the roots of x2-2qx+3=0
 So,β+γ=2q     
3βγ=3           
4And γ,α are the roots of x2-2rx+6=0
 So,γ+α=2r     
5γα=6           
6Multiply equation 246, we get,αβγ2=2×3×6⇒αβγ=6          
7Now divide equation246 by equation7, we get,γ=3, α=2 and β=1Adding equation135, we get,α+β+β+γ+γ+α=2p+2q+2r⇒2α+β+γ=2p+q+r⇒22+1+3=2p+q+r⇒p+q+r=6Hence Proved

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(α,β);(β,ϒ) and (ϒ,α) are respectively the roots of x2 -2px+2 =0, x2 -2qx +3=0 and x2 -2rx+6=0.If α,β and ϒ are all positive ,then prove that the value of p+q+r is 6

(α,β);(β,ϒ) and (ϒ,α) are respectively the roots of x² -2px+2 =0, x² -2qx +3=0 and x² -2rx+6=0.If α,β and ϒ are all positive ,then prove that the value of p+q+r is 6