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Question
If α,β are the roots of the equation ax2+bx+c=0 then the roots of the equation
(a+b+c)x2−(b+2c)x+c=0 are
If α,β are the roots of the equation ax2+bx+c=0 then the roots of the equation
(a+b+c)x2−(b+2c)x+c=0 are
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Solution
The correct option is D αα−1,ββ−1
If α,β, are the roots of the ............................α+β=−ba,αβ=fracca (a+b+c)x2−(b+2c)x+c=0(1+ba+ca)x2−(ba+2ca)x+ca=0(1−α−β+αβ)x2+(α+β−2αβ)=0x2+α+β−αβ−αβ1−α−β+αβx+αβ1−α−β+αβ=0x2+α(1−β)+β(1−α)(1−α)(1−beta)x+αβ(1−α)(1−beta)=0x2−[−α1−α−β1−β]x+(−α1−α)(β1−β)=0x2−(α1−α+β1−β)x+(αα−1)(ββ−1)=0⇒ root αα−1 and ββ−1Alternative :(a+b+c)x2−(b+2c)x+c=0ax2+b(x2−x)+c(x2−2x+1)=0a(xx−1)2+b(xx−1)+c=0Put t=xx−1at2+bt+c=0⇒t=α,t=β⇒xx−1=α,xx−1=βx=αα−1,x=ββ−1
If α,β, are the roots of the ............................
α+β=−ba,αβ=fracca
(a+b+c)x2−(b+2c)x+c=0
(1+ba+ca)x2−(ba+2ca)x+ca=0
(1−α−β+αβ)x2+(α+β−2αβ)=0
x2+α+β−αβ−αβ1−α−β+αβx+αβ1−α−β+αβ=0
x2+α(1−β)+β(1−α)(1−α)(1−beta)x+αβ(1−α)(1−beta)=0
x2−[−α1−α−β1−β]x+(−α1−α)(β1−β)=0
x2−(α1−α+β1−β)x+(αα−1)(ββ−1)=0
⇒ root αα−1 and ββ−1
Alternative :
(a+b+c)x2−(b+2c)x+c=0
ax2+b(x2−x)+c(x2−2x+1)=0
a(xx−1)2+b(xx−1)+c=0
Put t=xx−1
at2+bt+c=0
⇒t=α,t=β
⇒xx−1=α,xx−1=β
x=αα−1,x=ββ−1
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