Multiple choice

Find the value of 'k' for which the following set of quadratic equation has exactly one common root ${x^{2 }}-kx + 10 =0 $ and ${x^2} + kx - 18 = 0.$

  1. $ \pm 3$
  2. $ \pm 7$
  3. $ \pm 9$
  4. $ \pm 11$
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B Correct answer
Explanation

For a common root 'a', we have a^2 - ka + 10 = 0 and a^2 + ka - 18 = 0. Subtracting the equations gives 2ka = 28, so ka = 14. Substituting a = 14/k into the first equation gives (14/k)^2 - 14 + 10 = 0, so 196/k^2 = 4, which means k^2 = 49, so k = +/- 7.

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