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Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and the coefficients.

6x2 – 3 – 7x

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Given polynomial is 6x2– 3 – 7x 

We have, 6x2 – 3 – 7x = 6x2 – 7x – 3 

= 6x2 – 9x + 2x – 3 

= 3x(2x – 3) + 1(2x – 3) 

= (2x – 3) (3x + 1) 

The value of 6x2 – 3 – 7x is zero, when the value of (3x +1) (2x – 3) is 0 

i.e., when 3x + 1 = 0 and 2x – 3 = 0 

3x = -1 and 2x = 3

x = -1/3 and x = 3/2

∴ The zeroes of 6x2 – 3 – 7x = -1/3 and 3/2

∴ Sum of the zeroes = 1/3 + 3/2 = 7/6.

\(-\frac{Coefficient\,of\,x}{Coefficient\,of\,x^2}=\frac{-(-7)}{6}=\frac{7}{6}\)

And product of the zeroes = (-1/3) x (3/2) = -1/2

\(\frac{Constant\,term}{Coefficient\,of\,x^2}= \frac{-3}{6}=\frac{-1}{2}\)

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