Multiple choice

Find the value of $k$ for the following quadratic equation, so that they have two real and equal roots: $kx^2 - 2 \sqrt 5x + 4 = 0$

  1. $\displaystyle k = \frac{4}{5}$
  2. $\displaystyle k = -\frac{4}{5}$
  3. $\displaystyle k = \frac{5}{4}$
  4. $\displaystyle k = -\frac{5}{4}$
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C Correct answer
Explanation

For equal roots, D = b^2 - 4ac = 0. Here, (-2*sqrt(5))^2 - 4(k)(4) = 0. So 20 - 16k = 0, which means 16k = 20, or k = 20/16 = 5/4.

AI explanation

For a quadratic equation to have real and equal roots, its discriminant (b^2 - 4ac) must be zero. Here, a = k, b = -2 root 5, and c = 4, so setting the discriminant to zero gives (-2 root 5)^2 - 4(k)(4) = 0. This simplifies to 20 - 16k = 0. Solving for k gives k = 20 divided by 16, which is 5/4.

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