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Question
If one of the roots of the equation x2+ax+b=0 and x2+bx+a=0 is incident, then the numerical value of (a+b) is
If one of the roots of the equation x2+ax+b=0 and x2+bx+a=0 is incident, then the numerical value of (a+b) is
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Solution
The correct option is C −1
α+β=−a ;αβ=b
α+γ=−b ;αr=aa+b=−(2α+(β+r))=α(β+γ)..........(i)α+β=−αγ⇒α+β+αγ=0.............(ii)α+r=−αβ⇒α+γ+αβ=0.............(iii)(ii)−(iii)⇒(β−γ)(1−α)=0⇒α=1⇒1+β=−γ ⇒γ+β=−1
a+b=α(β+γ)=(1)(−1)=−1 Answer.
α+β=−a ;αβ=b
α+γ=−b ;αr=a
a+b=−(2α+(β+r))=α(β+γ)..........(i)
α+β=−αγ⇒α+β+αγ=0.............(ii)
α+r=−αβ⇒α+γ+αβ=0.............(iii)
(ii)−(iii)⇒(β−γ)(1−α)=0
⇒α=1⇒1+β=−γ
⇒γ+β=−1
a+b=α(β+γ)=(1)(−1)
a+b=α(β+γ)=(1)(−1)
=−1 Answer.
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