If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $f(x) = x^2 - p x + q$, form a quadratic polynomial whose zeroes are $\dfrac{\alpha^2}{\beta}$ and $\dfrac{\beta^2}{\alpha}$.
Step-by-Step Solution
Key Concept: $\alpha + \beta = p, \alpha \beta = q$.<br>New Sum $= \dfrac{\alpha^2}{\beta} + \dfrac{\beta^2}{\alpha} = \dfrac{\alpha^3 + \beta^3}{\alpha \beta} = \dfrac{p(p^2 - 3q)}{q}$.<br>New Product $= \dfrac{\alpha^2 \beta^2}{\alpha \beta} = \alpha \beta = q$.
$\alpha + \beta = p, \alpha \beta = q$. $\alpha^3 + \beta^3 = p(p^2 - 3q)$. [1.5 Marks]
New Sum $= \dfrac{\alpha^3 + \beta^3}{\alpha \beta} = \dfrac{p(p^2 - 3q)}{q} = \dfrac{p^3 - 3pq}{q}$. [1.5 Marks]
New Product $= \alpha \beta = q$. [1.0 Mark]
Required polynomial is $x^2 - \left(\dfrac{p^3 - 3pq}{q}\right)x + q$ or $q x^2 - p(p^2 - 3q)x + q^2$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating $\alpha^3 + \beta^3 = p(p^2 - 3q)$: 1.5 Marks
Calculating new sum and new product: 2.5 Marks
Writing polynomial $q x^2 - (p^3 - 3pq)x + q^2$: 1.0 Mark
Correct Answer: