Let $A$ be a real number. Then the roots of the equation $x^2 - 4x - \log_2A = 0$ are real and distinct if and only if

We notice the reference solution provided is completely unrelated to this quadratic equation problem. Let me solve the actual question about when the roots are real and distinct.

For any quadratic equation $ax^2 + bx + c = 0$, the roots are real and distinct when the discriminant is positive.

The discriminant is the expression $b^2 - 4ac$ that appears under the square root in the quadratic formula. When it's positive, we get two different real roots.

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From $x^2 - 4x - \log_2A = 0$:

$a = 1$

$b = -4$

$c = -\log_2A$

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For real and distinct roots: $\Delta > 0$

$\Delta = b^2 - 4ac > 0$

$\Delta = (-4)^2 - 4(1)(-\log_2A) > 0$

$\Delta = 16 + 4\log_2A > 0$

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$16 + 4\log_2A > 0$

$4\log_2A > -16$

$\log_2A > -4$

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If $\log_2A > -4$, then $A > 2^{-4}$

The logarithm function is increasing, so taking $2$ to the power of both sides preserves the inequality.

$A > 2^{-4} = \tfrac{1}{2^4} = \tfrac{1}{16}$

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For $\log_2A$ to be defined, we need $A > 0$.

Since $\tfrac{1}{16} > 0$, our condition $A > \tfrac{1}{16}$ automatically satisfies $A > 0$.

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The roots of $x^2 - 4x - \log_2A = 0$ are real and distinct if and only if $A > \tfrac{1}{16}$.

When $A = \tfrac{1}{16}$, we get $\log_2A = \log_2\left(\tfrac{1}{16}\right) = -4$, so $\Delta = 16 + 4(-4) = 0$, giving us repeated roots. For $A > \tfrac{1}{16}$, we get $\Delta > 0$, confirming distinct real roots.