Multiple choice

In the quadratic equation $\displaystyle ax^{2}+bx+c=0,a\neq 0,\Delta =b^{2}-4ac: : and: : \alpha +\beta ,\alpha ^{2}+\beta ^{2},\alpha ^{3}+\beta ^{3}$ are in GP where $\displaystyle \alpha ,\beta $ are the root of $\displaystyle ax^{2}+bx+c=0$ then

  1. $\displaystyle \Delta \neq 0$
  2. $\displaystyle b\Delta=0$
  3. $\displaystyle c\Delta=0$
  4. $\displaystyle \Delta=0$
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C Correct answer
Explanation

If a, b, c are in GP, then b^2 = ac. Here, (alpha^2 + beta^2)^2 = (alpha + beta)(alpha^3 + beta^3). Expanding and using symmetric sums of roots leads to the condition c*Delta = 0.

AI explanation

Using Vieta's formulas, the sum of the roots alpha plus beta equals negative b divided by a, and their product alpha beta equals c divided by a. The terms alpha plus beta, alpha squared plus beta squared, and alpha cubed plus beta cubed form a geometric progression, so the square of alpha squared plus beta squared equals the product of the other two terms. Substituting alpha squared plus beta squared equals b squared minus 2ac divided by a squared and alpha cubed plus beta cubed equals 3abc minus b cubed minus c squared times a divided by a cubed, then simplifying the resulting equation by factoring out c, yields c times b squared minus 4ac equals 0, which is c times delta equals 0.

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