Multiple choice

For what values of $k$ will the quadratic equation : $\displaystyle { 2x }^{ 2 }-kx+1=0$ have real and equal roots?

  1. $\displaystyle \pm 2\sqrt { 2 } $
  2. $\displaystyle \pm \sqrt { 2 } $
  3. $\displaystyle \pm 3\sqrt { 2 } $
  4. $\displaystyle \pm \sqrt { 3 } $
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A Correct answer
Explanation

For real and equal roots, D = b^2 - 4ac = 0. Here a=2, b=-k, c=1. D = (-k)^2 - 4(2)(1) = k^2 - 8 = 0. k^2 = 8, so k = +/- 2*sqrt(2).

AI explanation

For a quadratic equation to have real and equal roots, its discriminant must be zero. Applying the discriminant formula of D = b^2 - 4ac to 2x^2 - kx + 1 = 0, we get (-k)^2 - 4(2)(1) = 0. This simplifies to k^2 = 8, so taking the square root gives the value of k as plus or minus 2 times the square root of 2.

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