Multiple choice

If a root of the equation $ax^2+bx+c=0$ be reciprocal of a root of the equation $a'x^2+b'x+c'=0$, then?

  1. $(cc'-aa')^2=(ba'-cb')(ab'-bc')$
  2. $(bb'-aa')^2=(ca'-bc')(ab'-bc')$
  3. $(cc'-aa')^2=(ba'+cb')(ab'+bc')$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let alpha be a root of the first equation. Then 1/alpha is a root of the second. Substituting 1/alpha into the second equation and using the first equation's root property leads to the condition (cc'-aa')^2 = (ba'-cb')(ab'-bc').

AI explanation

Let the common root be r, so it is a root of the first equation ax^2 + bx + c = 0 and its reciprocal 1/r is a root of the second equation a'x^2 + b'x + c' = 0. Substituting these values gives the two equations ar^2 + br + c = 0 and c'r^2 + b'r + a' = 0. Using the cross-multiplication method to eliminate r, we equate the ratios of the coefficients, which results in the determinant formula (cc' - aa')^2 = (ba' - cb')(ab' - bc').

Keep practicing — related questions