Geometric Meaning of Zeroes
A real number $k$ is a zero of polynomial $p(x)$ if $p(k) = 0$.
- Geometrically, the zeroes of $y = p(x)$ are the $x$-coordinates of the points where the curve intersects the $X$-axis.
- A polynomial of degree $n$ can have at most $n$ real zeroes.
- A quadratic polynomial $p(x) = ax^2 + bx + c$ graph is a parabola (opens upwards if $a > 0$, downwards if $a < 0$).
Zeroes and Coefficients of a Quadratic Polynomial
If $\alpha$ and $\beta$ are the zeroes of quadratic polynomial $p(x) = ax^2 + bx + c$ ($a \neq 0$):
$$\text{Sum of Zeroes: } \alpha + \beta = -\frac{b}{a} = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$$
$$\text{Product of Zeroes: } \alpha \beta = \frac{c}{a} = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$$
Forming a Quadratic Polynomial with Given Zeroes:
$$p(x) = k [x^2 - (\alpha + \beta)x + \alpha\beta] \quad (k \neq 0)$$
Solved CBSE Board Problem
- Problem (3 Marks): Find the zeroes of the quadratic polynomial $6x^2 - 3 - 7x$ and verify the relationship between the zeroes and coefficients.
- Solution:
- Rewrite in standard form: $p(x) = 6x^2 - 7x - 3$.
- Splitting the middle term (product $= -18$, sum $= -7$):
$$6x^2 - 9x + 2x - 3 = 3x(2x - 3) + 1(2x - 3) = (2x - 3)(3x + 1)$$
- Setting $p(x) = 0$: $\alpha = \frac{3}{2}, \beta = -\frac{1}{3}$.
- Verification:
- Sum: $\alpha + \beta = \frac{3}{2} - \frac{1}{3} = \frac{7}{6}$. Here $-\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6}$ (Verified).
- Product: $\alpha\beta = \left(\frac{3}{2}\right)\left(-\frac{1}{3}\right) = -\frac{1}{2}$. Here $\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$ (Verified).
