Question

If the equation \( x^2 - 3ax + a^2 - 2a - K = 0 \) has different real roots for every rational number \( a \), then \( K \) lies in the interval

• A. \( 0<K<\frac{4}{5} \)
• B. \( -\infty<K<\frac{4}{5} \)
• C. \( \frac{4}{5}<K<\infty \)
• D. \( -\infty<K<\infty \)

Hint:

If a quadratic expression \( Ax^2 + Bx + C \) is always positive for all real \( x \), then \( A>0 \) and Discriminant \(<0 \).

Answer 1

The Correct Option is C

Step 1: Analysis of Roots:

For the quadratic equation in \( x \), \( x^2 - 3ax + (a^2 - 2a - K) = 0 \), to have different real roots, its discriminant \( \Delta_x \) must be strictly positive. \[ \Delta_x = (-3a)^2 - 4(1)(a^2 - 2a - K)>0 \] \[ 9a^2 - 4a^2 + 8a + 4K>0 \] \[ 5a^2 + 8a + 4K>0 \]
Step 2: Condition on 'a':

The problem states this must hold for every rational number \( a \). Since the rational numbers are dense in the real numbers and the function \( f(a) = 5a^2 + 8a + 4K \) is a continuous parabola, the condition implies \( f(a)>0 \) (or \( \ge 0 \) in limit, but "different" implies strict) for all real \( a \). Wait, technically if \( f(a)>0 \) for all \( a \in \mathbb{Q} \), by density/continuity it must be \( f(a) \ge 0 \) for all \( a \in \mathbb{R} \). However, if \( f(a) \) touches 0 at a rational point, the roots become equal, violating "different". If it touches 0 at an irrational point, for rational \( a \) it is still positive. However, usually in such problems, we ensure the minimum of the quadratic in \( a \) is strictly positive to be safe for all values. Since the coefficient of \( a^2 \) (which is 5) is positive, the parabola opens upward. For it to be positive for all \( a \), its minimum value must be positive. Alternatively, the discriminant of this quadratic in \( a \), let's call it \( \Delta_a \), must be negative (so it has no real roots and stays above the axis).
Step 3: Solving the inequality:

Method 1: Vertex/Minimum calculation. Minimum of \( 5a^2 + 8a + 4K \) occurs at \( a = \frac{-8}{10} = -\frac{4}{5} \). \[ f_{\min} = 5\left(-\frac{4}{5}\right)^2 + 8\left(-\frac{4}{5}\right) + 4K \] \[ = 5\left(\frac{16}{25}\right) - \frac{32}{5} + 4K \] \[ = \frac{16}{5} - \frac{32}{5} + 4K = -\frac{16}{5} + 4K \] We require \( f_{\min}>0 \): \[ 4K - \frac{16}{5}>0 \] \[ 4K>\frac{16}{5} \implies K>\frac{4}{5} \] Method 2: Discriminant of \( f(a) \)<0. \( \Delta_a = 8^2 - 4(5)(4K)<0 \) \( 64 - 80K<0 \) \( 64<80K \) \( K>\frac{64}{80} = \frac{4}{5} \)
Step 4: Final Answer:

\( K \) lies in the interval \( \left(\frac{4}{5}, \infty\right) \).