Find the zeroes of the quadratic polynomial $p(x) = \sqrt{3} x^2 - 8x + 4\sqrt{3}$ and verify relationship with coefficients.
Step-by-Step Solution
Key Concept: Split $-8x = -6x - 2x$ since $\sqrt{3} \times 4\sqrt{3} = 12$.
$\sqrt{3}x^2 - 6x - 2x + 4\sqrt{3} = 0 \Rightarrow \sqrt{3}x(x - 2\sqrt{3}) - 2(x - 2\sqrt{3}) = 0$.
$(\sqrt{3}x - 2)(x - 2\sqrt{3}) = 0 \Rightarrow x = 2/\sqrt{3} = 2\sqrt{3}/3$ and $x = 2\sqrt{3}$. [1.5 Marks]
Sum $= 2/\sqrt{3} + 2\sqrt{3} = 8/\sqrt{3} = -(-8)/\sqrt{3} = -b/a$.
Product $= (2/\sqrt{3})(2\sqrt{3}) = 4 = 4\sqrt{3}/\sqrt{3} = c/a$. Verified! [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Finding zeroes $2/\sqrt{3}$ and $2\sqrt{3}$: 1.5 Marks
Verifying sum and product relationships: 1.5 Marks
Correct Answer: