Find the zeroes of the quadratic polynomial $x^2 + 7x + 10$, and verify the relationship between the zeroes and the coefficients.
Step-by-Step Solution
Key Concept: Factorise $p(x) = (x+2)(x+5) = 0$ to get zeroes $\alpha, \beta$, then check $\alpha + \beta = -b/a$ and $\alpha \beta = c/a$.
$p(x) = x^2 + 7x + 10 = x^2 + 5x + 2x + 10 = x(x+5) + 2(x+5) = (x+2)(x+5)$. Zeroes are $\alpha = -2$ and $\beta = -5$. [1.0 Mark]
Verification:
Sum of zeroes $= \alpha + \beta = -2 + (-5) = -7 = -\dfrac{b}{a} = -\dfrac{7}{1}$. Verified!
Product of zeroes $= \alpha \beta = (-2)(-5) = 10 = \dfrac{c}{a} = \dfrac{10}{1}$. Verified! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Finding zeroes (-2 and -5) by factorisation: 1.0 Mark
Verifying sum and product relationships with coefficients: 1.0 Mark
Correct Answer: