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If there is only one positive real solution `x` to the equation `(x^2+kx+3)/(x-1)=3x+k`...
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First, it is impossible to have only one real solution to a quadratic equation. There exists either two real solutions, distinct or equal, or two complex conjugate solutions. Since the problem provides the information that the equation has only one real positive solution, then the other solution either is real negative or it is equal to the positive solution.
Considering the other real root negative, thus the product of the roots is also negative. Using Vieta's relations yields:
`x_1*x_2 = c/a < 0`
Identifying `a = 2` and `c = -k-3` yields:
`-(k+3)/2 < 0 => -(k+3) < 0 => k+3 > 0 => k > -3`
If the roots are both positive, then `Delta = 0 => 9 + 8(k + 3) = 0 => 8k + 24 + 9 = 0 => 8k = -33 => k = -33/8`
Hence, the conditions you need to consider, using the given information, are `k > -3` (one positive solution and one negative solution) or `k = -33/8` (two equal positive solutions).
Posted by sciencesolve on July 12, 2013 at 11:02 AM (Answer #2)
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