Multiple choice

If rational numbers $a,b,c,d$ are in G.P., then roots of equation $\displaystyle (a-c)^{2}x^{2}+(b-c)^{2}x+(b-d)^{2}= (a-d)^{2} $ are necessarily

  1. Imaginary

  2. Irrational

  3. rational

  4. real and distinct
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If a, b, c, d are in G.P., let b=ar, c=ar^2, d=ar^3. Substituting these into the equation (a-c)^2 x^2 + (b-c)^2 x + (b-d)^2 = (a-d)^2 and simplifying shows the coefficients are rational expressions of a and r. Since a, b, c, d are rational, the coefficients are rational, leading to rational roots.

AI explanation

Because a, b, c, d are in a geometric progression, we can substitute b = ar, c = ar^2, and d = ar^3. The equation becomes (a - ar^2)^2*x^2 + (ar - ar^2)^2*x + (ar - ar^3)^2 = (a - ar^3)^2. Factoring out a^2 gives a^2(1 - r^2)^2*x^2 + a^2*r^2(1 - r)^2*x + a^2*r^2(1 - r^2)^2 = a^2(1 - r^3)^2. We observe that one root is exactly 1, since substituting x = 1 makes the left side a perfect square that equals the right side: (a - ar^2 + ar - ar^2)^2 = (a - ar^3)^2. Since a, b, c, d are rational numbers and r is rational, all coefficients of the quadratic equation are rational. A quadratic equation with rational coefficients that has one rational root must necessarily have both roots as rational numbers.

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