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Question
If α, β are real and α2, −β2 are the roots of the equations a2x2+x+(1−a2)=0, a>1, then β2=
If α, β are real and α2, −β2 are the roots of the equations a2x2+x+(1−a2)=0, a>1, then β2=
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Solution
The correct option is B 1
We have,a2x2+x+1−a2=0a2(x2−1)+(x+1)=0a2(x−1)(x+1)+(x+1)=0(x+1)[a2(x−1)+1]=0x=−1a2x=a2−1x=1−1a2−β2=−1β2=1
Option B is correct.
We have,
a2x2+x+1−a2=0a2(x2−1)+(x+1)=0a2(x−1)(x+1)+(x+1)=0(x+1)[a2(x−1)+1]=0x=−1a2x=a2−1x=1−1a2−β2=−1β2=1
Option B is correct.
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