Multiple choice

The ratio of the roots of the equation $a{ x }^{ 2 }+bx+c=0$ is same as the ratio of the roots of the equation $p{ x }^{ 2 }+qx+r=0$. If ${ D }{ 1 }$ and ${ D }{ 2 }$ are the discriminants of $a{ x }^{ 2 }+bx+c=0$ and $p{ x }^{ 2 }+qx+r=0$ respectively, then ${ D }{ 1 }:{ D }{ 2 }$ is equal to

  1. $\displaystyle \frac { { a }^{ 2 } }{ { p }^{ 2 } } $
  2. $\displaystyle \frac { { b }^{ 2 } }{ { q }^{ 2 } } $
  3. $\displaystyle \frac { { c }^{ 2 } }{ { r }^{ 2 } } $

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

Let the roots be alpha, beta. Ratio alpha/beta = k. For ax^2+bx+c=0, roots are (-b +/- sqrt(D1))/2a. Ratio = (-b+sqrt(D1))/(-b-sqrt(D1)) = k. This implies D1/b^2 is constant. Specifically, D1/b^2 = (k-1)^2 / (k+1)^2. Since this ratio is the same for both equations, D1/b^2 = D2/q^2, so D1/D2 = b^2/q^2.

AI explanation

Let the ratio of the roots for both equations be k:1. By Vieta's formulas for ax^2 + bx + c = 0, if the roots are ky and y, their sum is y(k+1) = -b/a and product is ky^2 = c/a. Solving for the discriminant D1 = b^2 - 4ac, we substitute b = -ay(k+1) and c = aky^2 to get D1 = a^2*y^2*(k+1)^2 - 4*a^2*ky^2 = a^2*y^2*(k-1)^2. Applying the same method to px^2 + qx + r = 0 gives D2 = p^2*y^2*(k-1)^2. The ratio D1/D2 simplifies to a^2/p^2, and since b/q = a/p, this equals b^2/q^2.

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