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Polynomials
RD Sharma
CBSE
Grade 10

Question:

If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial $p(x) = 2x^2 + 5x + k$ such that $\alpha^2 + \beta^2 + \alpha \beta = \dfrac{21}{4}$, find the value of $k$. Hence, find the actual zeroes of the polynomial.

Step-by-Step Solution

Key Concept: $k = 2$. $p(x) = 2x^2 + 5x + 2 = (2x + 1)(x + 2) = 0 \Rightarrow$ zeroes are $-1/2$ and $-2$.
$\alpha + \beta = -5/2, \alpha \beta = k/2$. $(\alpha + \beta)^2 - \alpha \beta = 25/4 - k/2 = 21/4 \Rightarrow k/2 = 1 \Rightarrow k = 2$. [2.5 Marks]
Polynomial becomes $2x^2 + 5x + 2 = (2x + 1)(x + 2) = 0$. [1.5 Marks]
Zeroes are $x = -1/2$ and $x = -2$. [1.0 Mark]

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🎯 Official CBSE Marking Scheme:
Solving $k = 2$: 2.5 Marks
Factoring $2x^2 + 5x + 2$: 1.5 Marks
Finding zeroes $-1/2$ and $-2$: 1.0 Mark

Correct Answer:
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