Question

Let A be a real number. Then the roots of the equation \(x^2-4x-log_{2}A=0 \) are real and distinct if and only if

• A. \(A>\frac{1}{16}\)
• B. \(A>\frac{1}{8}\)
• C. \(A<\frac{1}{16}\)
• D. \(A<\frac{1}{8}\)

Answer 1

The Correct Option is A

To determine when the roots of the quadratic equation \(x^2-4x-\log_{2}A=0\) are real and distinct, we need to analyze the discriminant of the quadratic formula. The quadratic equation is of the form \(ax^2+bx+c=0\), where \(a=1\), \(b=-4\), and \(c=-\log_{2}A\).

The discriminant \(\Delta\) for a quadratic equation \(ax^2+bx+c=0\) is given by the formula:

\[\Delta = b^2-4ac\] 

Substituting the values, we have:

\[\Delta = (-4)^2-4(1)(-\log_{2}A)\]

\[\Delta = 16+4\log_{2}A\]

For the roots to be real and distinct, the discriminant must be greater than zero:

\[16+4\log_{2}A>0\]

This simplifies to:

\[4\log_{2}A>-16\]

Dividing both sides by 4, we get:

\[\log_{2}A>-4\]

Converting from logarithmic form to exponential form, we have:

\[A>2^{-4}\]

Simplifying further, we find:

\[A>\frac{1}{16}\]

Therefore, the roots of the equation will be real and distinct if \(A>\frac{1}{16}\).

Option	Condition
A	\(A>\frac{1}{16}\)
B	\(A>\frac{1}{8}\)
C	\(A<\frac{1}{16}\)
D	\(A<\frac{1}{8}\)

Thus, the correct condition for the roots to be real and distinct is \(A>\frac{1}{16}\).