Mathematics — CBSE Class 10 board question

For $p(y) = y^2 - 4\sqrt{3}y + 3$, comparing with $ay^2 + by + c$:
$a = 1,\ b = -4\sqrt{3},\ c = 3$

Using Vieta's formulas:
$$\alpha + \beta = \frac{-b}{a} = 4\sqrt{3}, \qquad \alpha\beta = \frac{c}{a} = 3$$

Therefore:
$$4\sqrt{3} - 3\cdot4 = (\alpha+\beta) - 4(\alpha\beta) = 4\sqrt{3} - 4(3) = 4\sqrt{3} - 12$$

Note: The expression $4\sqrt{3} - 3\cdot4$ evaluates numerically as $4\sqrt{3} - 12 \approx 6.93 - 12 = -5.07$, but in terms of the polynomial's coefficients, the answer is $\boxed{4\sqrt{3} - 12}$.

Source: Chapter 2, Section 2.3 – Relationship between Zeroes and Coefficients of a Polynomial

---

• The question asks to find $4\sqrt{3} - 3 \times 4$, which maps directly to $(\alpha+\beta) - 4(\alpha\beta)$ using the sum and product of zeroes.
• Examiners expect you to first state the formulas, substitute $a, b, c$ correctly, then evaluate the given expression by substitution.
• The key skill tested is recognising that $4\sqrt{3} = \alpha+\beta$ and $3 = \alpha\beta$ from the polynomial's coefficients, so the expression becomes $4\sqrt{3} - 4(3) = 4\sqrt{3} - 12$.