Multiple choice

Find the value of $k$ for which given equation has real and equal roots. $(k\,-\,12)\,x^2\,+\,2\,(k\,-\,12)\,x\,+\,2\,=\,0$

  1. $k\,=\,12$
  2. $k\,=\,13$
  3. $k\,=\,14$
  4. $k\,=\,15$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For real and equal roots, discriminant D = b^2 - 4ac = 0. Here a = (k-12), b = 2(k-12), c = 2. D = [2(k-12)]^2 - 4(k-12)(2) = 4(k-12)^2 - 8(k-12) = 0. Divide by 4(k-12): (k-12) - 2 = 0, so k = 14.

AI explanation

For the given equation to have real and equal roots, its discriminant must be equal to zero, so b^2 - 4ac = 0. Substituting a = (k - 12), b = 2(k - 12), and c = 2 gives the equation (2(k - 12))^2 - 4(k - 12)(2) = 0. Factoring out 4(k - 12) results in 4(k - 12)((k - 12) - 2) = 0, which means k = 14 because k cannot equal 12 in a quadratic equation.

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