Multiple choice

Let $\displaystyle \alpha ,\beta $ be the roots of the equation $\displaystyle x^{2}-px+r=0$ and $\displaystyle {\alpha }/{2},2\beta$ be the roots of the equation $\displaystyle x^{2}-qx+r=0.$ Then the value of $r$ is

  1. $\displaystyle {2}/{9}\left ( p-q \right )\left ( 2q-p \right )$
  2. $\displaystyle {2}/{9}\left ( q-p \right )\left ( 2p-q \right )$
  3. $\displaystyle {2}/{9}\left ( q-2p \right )\left ( 2q-p \right )$
  4. $\displaystyle {2}/{9}\left ( 2p-q \right )\left ( 2q-p \right )$
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D Correct answer
Explanation

For x^2 - px + r = 0, alpha + beta = p, alpha * beta = r. For x^2 - qx + r = 0, alpha/2 + 2beta = q, (alpha/2) * 2beta = r. So alpha * beta = r. From alpha/2 + 2beta = q, substitute beta = r/alpha: alpha/2 + 2r/alpha = q. alpha^2 - 2q*alpha + 4r = 0. Using roots of first eq, alpha^2 - p*alpha + r = 0. Subtracting: (p-2q)alpha = 3r. alpha = 3r/(p-2q). Substitute back to get r.

AI explanation

Using Vieta's formulas, the sum and product of the roots for the first equation are alpha + beta = p and alpha * beta = r. For the second equation, the sum of the roots is (alpha / 2) + 2 * beta = q. Multiplying this sum by 2 gives alpha + 4 * beta = 2q, and subtracting the first sum yields 3 * beta = 2q - p, so beta = (2q - p) / 3. We can similarly find alpha = (2p - q) / 3 by subtracting equations. Since r = alpha * beta, substituting the values gives r = (2p - q)(2q - p) / 9, which equals 2/9 * (2p - q)(2q - p).

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