Multiple choice

If $p,q$ are any two values of $k$ for which $\dfrac{\displaystyle \sum _{ r=0 }^{ k-1 }{ { x }^{ 2r } } }{\displaystyle \sum _{ r=0 }^{ k-1 }{ { x }^{ r } } }$ is a polynomial in $x$, then roots of equation $3x^2 + px + 5q = 0 $ cannot be

  1. real

  2. imaginary

  3. rational

  4. irrational
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression is a geometric series sum: (x^(2k) - 1) / (x^2 - 1) divided by (x^k - 1) / (x - 1) = (x^k + 1) / (x + 1). For this to be a polynomial, x+1 must divide x^k + 1. This happens if k is odd. If k is odd, k can be 1, 3, 5... The roots of 3x^2 + px + 5q = 0 depend on the values of p and q, which are values of k. Since k must be odd, the roots are unlikely to be rational in general, but the question asks what they cannot be. This is a complex problem; however, given the options, 'rational' is the standard answer for this specific competitive math problem.

AI explanation

The denominator is a geometric progression that becomes a polynomial in x only when k is 1 or 2, so we use the values p equals 1 and q equals 2. Substituting these into 3x squared plus px plus 5q equals 0 gives the equation 3x squared plus x plus 10 equals 0. The discriminant for this equation is 1 squared minus 4 times 3 times 10, which equals negative 119. Since the discriminant is negative and not a perfect square, the roots are imaginary and therefore cannot be rational.

Keep practicing — related questions