Multiple choice

Given $\alpha $ and $\beta $ are the roots of the equation ${x^2} - 4x + k = 0(k \ne 0)$. If $\alpha \beta ,\ \alpha {\beta ^2} + {\alpha ^2}\beta ,\ {\alpha ^3} + {\beta ^3}$ are in geometric progression then the value of $'k'$ equals

  1. $4$
  2. $\dfrac {16}{7}$
  3. $\dfrac {3}{7}$
  4. $12$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Roots alpha, beta satisfy alpha+beta=4, alpha*beta=k. The terms are k, alpha*beta(alpha+beta)=4k, alpha^3+beta^3=(alpha+beta)((alpha+beta)^2-3alpha*beta)=4(16-3k). For GP, (4k)^2 = k * 4(16-3k). 16k^2 = 64k - 12k^2. 28k^2 = 64k. Since k is not 0, k = 64/28 = 16/7.

AI explanation

Using Vieta's formulas for the equation x^2-4x+k=0, the sum alpha+beta is 4 and the product alpha*beta is k. The terms in the geometric progression simplify to k, k(alpha+beta), and (alpha+beta)^3-3k(alpha+beta). Substituting the sum gives the progression as k, 4k, and 64-12k. For these to be in geometric progression, (4k)^2 equals k(64-12k). Solving 16k = 52-12k yields k = 16/7.

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