InterviewSolution
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InterviewSolution
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If the equations `x^2+bx-1=0` and `x^2+x+b=0` have a common root different from `-1` then `|b|` is equal toA. `sqrt(2)`B. 2C. `sqrt(3)`D. 3 |
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Answer» Correct Answer - C Let `alpha` a common root different from -1. Then, `alpha^(2) + b alpha - 1 = 0 and alpha^(2) + alpha + b = 0` `rArr" "(alpha^(2) + b alpha -1)-(alpha^(2) + alpha + b)=0 rArr alpha(b-1) = b+1 rArr alpha = (b+1)/(b-1)` Putting this value of `alpha "in" alpha^(2) + b alpha - 1 = 0`, we obtain `(b+1)^(2) + b(b^(2) -1) - (b-1)^(2) = 0 rArr b^(3) + 3b - 0 rArr b = 0, b = +- i sqrt(3)` For b = 0, `alpha^(2) + b alpha - 1 = 0` gives `alpha = - 1`. So b `ne` 0. `therefore" "b = +-i sqrt(3) rArr |b| = sqrt(3)` |
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