Multiple choice

If $r$ be the ratio of the roots of the equation $a{ x }^{ 2 }+bx+c=0$, then $\displaystyle \frac { { \left( r+1 \right) }^{ 2 } }{ r } =$

  1. $\displaystyle \frac { { a }^{ 2 } }{ bc } $
  2. $\displaystyle \frac { { b }^{ 2 } }{ ca } $
  3. $\displaystyle \frac { { c }^{ 2 } }{ ab } $

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots are a, ar. Sum = a(1+r) = -b/a. Product = a^2*r = c/a. (r+1)^2 / r = (r^2+2r+1)/r = r + 2 + 1/r = (r^2+1)/r + 2. Using sum/product: a(1+r) = -b/a => 1+r = -b/a^2. This approach is complex; standard identity is (r+1)^2/r = (sum of roots)^2 / (product of roots) = (-b/a)^2 / (c/a) = b^2 / (ac).

AI explanation

Let the roots of the equation be p and q. By Vieta's formulas, the sum of the roots is p + q = -b/a and the product is pq = c/a. Since r is the ratio of the roots, we can write p = rq. Substituting this into the sum gives rq + q = q(r + 1) = -b/a, and squaring this equation yields q^2(r + 1)^2 = b^2/a^2. Since q^2 equals pq divided by r, we can substitute q^2 = (c/a)/r to get (c/a)*(r + 1)^2/r = b^2/a^2. Multiplying both sides by a^2 and dividing by c gives (r + 1)^2/r = b^2/ca.

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