Polynomials Question - CBSE

Question

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = x^2 - x - 4$, find the value of:
(i) $\dfrac{1}{\alpha} + \dfrac{1}{\beta} - \alpha \beta$ [1.5 Marks]
(ii) $\alpha^4 + \beta^4$ [2.0 Marks]
(iii) $\alpha^2 \beta + \alpha \beta^2$ [1.5 Marks]

Accepted Answer

(i) $\dfrac{\alpha + \beta}{\alpha \beta} - \alpha \beta = \dfrac{1}{-4} - (-4) = -\dfrac{1}{4} + 4 = \dfrac{15}{4}$. [1.5 Marks]
(ii) $\alpha^2 + \beta^2 = 1^2 - 2(-4) = 9$.
$\alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 - 2\alpha^2 \beta^2 = 9^2 - 2(-4)^2 = 81 - 32 = 49$. [2.0 Marks]
(iii) $\alpha \beta (\alpha + \beta) = -4(1) = -4$. [1.5 Marks]

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🎯 Official CBSE Marking Scheme:
Part (i) Evaluating $15/4$: 1.5 Marks
Part (ii) Evaluating $\alpha^4 + \beta^4 = 49$: 2.0 Marks
Part (iii) Evaluating $-4$: 1.5 Marks

Question:

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