Question

If $ \frac{k}{kx + 3} + \frac{3}{3x-k}= \frac{12x + 5}{(kx + 3)(3x - k)} $, then both the roots of the equation $ kx^2 - 7x + 3 = 0 $ are:

• A. Rational numbers
• B. Irrational numbers
• C. Complex numbers
• D. Integers

Hint:

To determine the nature of roots, use the discriminant \( \Delta = b^2 - 4ac \). If \( \Delta \) is a perfect square and coefficients are rational, the roots are rational. Substitute values if necessary to test feasibility.

Answer 1

The Correct Option is A

Step 1: Combine the left-hand side using a common denominator. Given: \[ \frac{k}{kx + 3} + \frac{3}{3x - k} = \frac{12x + 5}{(kx + 3)(3x - k)} \] Take LHS: \[ \frac{k(3x - k) + 3(kx + 3)}{(kx + 3)(3x - k)} = \frac{12x + 5}{(kx + 3)(3x - k)} \] Step 2: Expand the numerator on the left-hand side. \[ k(3x - k) = 3kx - k^2 \] \[ 3(kx + 3) = 3kx + 9 \] Add them: \[ (3kx - k^2 + 3kx + 9) = 6kx - k^2 + 9 \] So, \[ \frac{6kx - k^2 + 9}{(kx + 3)(3x - k)} = \frac{12x + 5}{(kx + 3)(3x - k)} \] Step 3: Equate numerators. \[ 6kx - k^2 + 9 = 12x + 5 \] Group like terms: \[ 6kx - 12x = k^2 - 4 \Rightarrow x(6k - 12) = k^2 - 4 \] Solve for \( x \): \[ x = \frac{k^2 - 4}{6k - 12} \] This is a rational expression in terms of \( k \), so for the quadratic equation \( kx^2 - 7x + 3 = 0 \) to have rational roots, the discriminant must be a perfect square.
Step 4: Use discriminant condition for rational roots. The discriminant of \( kx^2 - 7x + 3 \) is: \[ \Delta = (-7)^2 - 4(k)(3) = 49 - 12k \] For roots to be rational, \( \Delta \) must be a perfect square. Try small integer values of \( k \) that satisfy earlier equation: Try \( k = 1 \): \[ x = \frac{1 - 4}{6 - 12} = \frac{-3}{-6} = \frac{1}{2} \Rightarrow \text{One rational value found, try in quadratic} \] Check discriminant: \[ \Delta = 49 - 12(1) = 37 \quad \text{(not a perfect square)} \] Try \( k = 2 \): \[ x = \frac{4 - 4}{12 - 12} = \frac{0}{0} \Rightarrow \text{Not defined} \] Try \( k = 4 \): \[ x = \frac{16 - 4}{24 - 12} = \frac{12}{12} = 1 \] Discriminant: \[ \Delta = 49 - 12(4) = 49 - 48 = 1 \quad \text{(perfect square)} \Rightarrow \text{Rational roots} \]