Question
If α,βare the roots of the equation ax2+bx+c=0 then the equation whose roots are α+1β and β+1α, is
Answer: Option A
:
A
Hereα+β= −baandαβ=ca
If roots are α+1β,β+1α, then sum of roots are
=(α+1β)+(β+1alpha)=(α+β)+α+βαβ=−bac(a+c)
and product =(α+1β)(β+1α)
=αβ+1+1+1αβ=2+ca+ac
=2ac+c2+a2ac=(a+c)2ac
Hence required equation is given by
x2+bac(a+c)x+(a+c)2ac=0
⇒ acx2+(a+c)bx+(a+c)2=0
Trick : Let a = 1, b = -3, c = 2,then α = 1,β = 2
∴α+1β=32andβ+1α=3
∴ required equation must be
(x−3)(2x−3)=0 i.e. 2x2−9x+9=0
Here (a) gives this equation on putting
a = 1, b = -3, c = 2
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:
A
Hereα+β= −baandαβ=ca
If roots are α+1β,β+1α, then sum of roots are
=(α+1β)+(β+1alpha)=(α+β)+α+βαβ=−bac(a+c)
and product =(α+1β)(β+1α)
=αβ+1+1+1αβ=2+ca+ac
=2ac+c2+a2ac=(a+c)2ac
Hence required equation is given by
x2+bac(a+c)x+(a+c)2ac=0
⇒ acx2+(a+c)bx+(a+c)2=0
Trick : Let a = 1, b = -3, c = 2,then α = 1,β = 2
∴α+1β=32andβ+1α=3
∴ required equation must be
(x−3)(2x−3)=0 i.e. 2x2−9x+9=0
Here (a) gives this equation on putting
a = 1, b = -3, c = 2
Was this answer helpful ?
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