Multiple choice

If the roots of the quadratic equation $x^2-2x-\log_{2}A=0$ are real, then the least value of $A$ is

  1. $1$
  2. $\dfrac{1}{2}$
  3. $\dfrac{1}{4}$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For real roots, the discriminant D = b^2 - 4ac >= 0. Here, (-2)^2 - 4(1)(-log2(A)) >= 0, so 4 + 4log2(A) >= 0. This implies log2(A) >= -1, so A >= 2^-1 = 1/2.

AI explanation

For the roots of x^2 - 2x - log2(A) = 0 to be real, the discriminant must be greater than or equal to zero, so (-2)^2 - 4(1)(-log2(A)) >= 0. This simplifies to 4 + 4log2(A) >= 0, or log2(A) >= -1. Converting this logarithmic inequality to exponential form gives A >= 2^(-1), meaning the least value of A is 1/2.

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