Find the zeroes of the polynomial $f(x) = 6x^2 - 3 - 7x$ and verify the relationship between the zeroes and the coefficients.
Step-by-Step Solution
Key Concept: Rearrange to standard form $6x^2 - 7x - 3 = 0$, factorise to find zeroes, then verify $\alpha+\beta = -b/a$ and $\alpha\beta = c/a$.
Rearranging: $f(x) = 6x^2 - 7x - 3 = 6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1)$. [1.0 Mark]
Zeroes are obtained when $2x - 3 = 0$ or $3x + 1 = 0 \Rightarrow \alpha = \dfrac{3}{2}$ and $\beta = -\dfrac{1}{3}$. [1.0 Mark]
Verification:
$\alpha + \beta = \dfrac{3}{2} + \left(-\dfrac{1}{3}\right) = \dfrac{9 - 2}{6} = \dfrac{7}{6} = -\dfrac{b}{a} = -\dfrac{-7}{6} = \dfrac{7}{6}$. Verified!
$\alpha \beta = \left(\dfrac{3}{2}\right)\left(-\dfrac{1}{3}\right) = -\dfrac{3}{6} = -\dfrac{1}{2} = \dfrac{c}{a} = \dfrac{-3}{6} = -\dfrac{1}{2}$. Verified! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Rearranging and factorising polynomial: 1.0 Mark
Finding zeroes ($3/2$ and $-1/3$): 1.0 Mark
Complete verification of sum and product relationships: 1.0 Mark
Correct Answer: