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Find the value of `k`, so that the equation `2x^2+kx-5=0` and `x^2-3x-4=0` may have one root in common.A. `-3, (27)/(4)`B. `3, (-27)/(4)`C. `-3, (-27)/(4)`D. `3, (27)/(4)` |
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Answer» Correct Answer - C Let `alpha` be the common root of the two equations. Then, `2alpha^(2) + k alpha - 5 = 0 and, alpha^(2) - 3alpha -4 = 0` Solving these two equations, we get `(alpha^(2))/(4k -15k)=(alpha)/(-5+8)=(1)/(-6-k)` `rArr" "alpha^(2) = (4k + 15)/(k+6) and alpha=(-3)/(k+6)` `rArr" "((-3)/(k+6))^(2) = (4k + 15)/(k+6)" "[therefore alpha^(2) = (alpha)^(2)]` `rArr" "(4k + 15) (k + 6) = 9` `rArr" "4k^(2) + 39k + 81 = 0 rArr k = - 3 or, k = -(27)/(4)` |
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