Multiple choice

The condition that a roots of the equation $ax^{2}+bx+c=0$ may be reciprocal to the roots of $a_{1}x^{2}+b_{1}x +c_{1}=0$ is

  1. $(bb_{1} -aa_{1})^{2}=(ab_{1}-bc_{1})(ba_{1}-b_{1}c)$
  2. $(cb_{1}-ba_{1})^{2}=(ac_{1}-cd_{1})(ab_{1}-bc_{1})$
  3. $(cc_{1}-aa_{1})^{2}=(ab_{1}-bc_{1})(ba_{1}-b_{1}c)$
  4. ${a}+{b}+{c}=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A root reciprocal to a root of the second equation is a common root of ax^2+bx+c and c1x^2+b1x+a1. Their resultant must be zero, which gives (cc1-aa1)^2 = (ab1-bc1)(ba1-b1c). This is the condition shown in option C.

AI explanation

Let the roots of the first equation be alpha and beta, and the roots of the second equation be 1 over alpha and 1 over beta. Using the relationships between roots and coefficients, we substitute a1 times alpha squared + b1 times alpha + c1 = 0 and a1 times beta squared + b1 times beta + c1 = 0. Eliminating the variables alpha and beta from these equations along with a times alpha squared + b times alpha + c = 0 yields the condition (cc1 - aa1) squared = (ab1 - bc1)(ba1 - b1c).

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