NCERT · Class 10 · Mathematics · PolynomialsFind the zeroes of the quadratic polynomial $f(x) = 6x^2 - 3 - 7x$ and verify the relationship between the zeroes and its coefficients.

Given the quadratic polynomial: $f(x) = 6x^2 - 7x - 3$

Step 1: Find the zeroes of the polynomial\nTo find the zeroes, set $f(x) = 0$: $6x^2 - 7x - 3 = 0$ \nSplit the middle term $-7x$ into two parts whose product is $(6)(-3) = -18$ and whose sum is $-7$. These numbers are $-9x$ and $+2x$: $6x^2 - 9x + 2x - 3 = 0$ \nFactor by grouping: $3x(2x - 3) + 1(2x - 3) = 0$ $(3x + 1)(2x - 3) = 0$ \nSetting each factor to zero: $3x + 1 = 0 \implies x = -\frac{1}{3}$ $2x - 3 = 0 \implies x = \frac{3}{2}$ \nThus, the zeroes of the polynomial are $\alpha = -\frac{1}{3}$ and $\beta = \frac{3}{2}$.

Step 2: Verify the relationship between zeroes and coefficients\nFor a standard quadratic polynomial $ax^2 + bx + c$, where $a=6, b=-7, c=-3$:

1. Sum of zeroes: $\alpha + \beta = -\frac{1}{3} + \frac{3}{2} = \frac{-2 + 9}{6} = \frac{7}{6}$ Using coefficients formula: $-\frac{b}{a} = -\frac{-7}{6} = \frac{7}{6}$ Since $\frac{7}{6} = \frac{7}{6}$, the sum of zeroes relationship is verified.

2. Product of zeroes: $\alpha \cdot \beta = \left(-\frac{1}{3}\right) \cdot \left(\frac{3}{2}\right) = -\frac{3}{6} = -\frac{1}{2}$ Using coefficients formula: $\frac{c}{a} = \frac{-3}{6} = -\frac{1}{2}$ Since $-\frac{1}{2} = -\frac{1}{2}$, the product of zeroes relationship is verified.