Multiple choice

The values of $k$, so that the equations $2x^{2} + kx - 5 = 0$ and $x^{2} - 3x - 4 = 0$ have one root in common, are

  1. $3, \dfrac {27}{2}$
  2. $9, \dfrac {27}{4}$
  3. $-3, \dfrac {-27}{4}$
  4. $-3, \dfrac {4}{27}$
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C Correct answer
Explanation

The roots of x^2 - 3x - 4 = 0 are (x-4)(x+1) = 0, so x = 4 or x = -1. If x = 4 is a common root, 2(4)^2 + k(4) - 5 = 0 => 32 + 4k - 5 = 0 => 4k = -27 => k = -27/4. If x = -1 is a common root, 2(-1)^2 + k(-1) - 5 = 0 => 2 - k - 5 = 0 => k = -3.

AI explanation

By factoring the second equation, x^2 - 3x - 4 = 0 = (x - 4)(x + 1), we find its roots are 4 and -1, meaning one of these must be the common root for 2x^2 + kx - 5 = 0. Substituting x = 4 into the first equation gives 2(16) + 4k - 5 = 0, which simplifies to 4k = -27 and k = -27/4. Substituting x = -1 into the first equation gives 2(1) - k - 5 = 0, which results in k = -3. The values of k are -3 and -27/4.

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