Find the zeroes of the quadratic polynomial $6x^2-3-7x$ and verify the relationship between the zeroes and the coefficients of the polynomial.

Given: The quadratic polynomial $6x^2-3-7x$.

To do: To find the zeroes of the quadratic polynomial $6x^2-3-7x$ and verify the relationship between the zeroes and the coefficients of the polynomial. 

Solution:

Given polynomial is $6x^2-3-7x$.

$=6x^2-7x-3$

$=6x^2-9x+2x-3$

$=3x( 2x-3)+1( 2x-3)$

$=( 2x-3)( 3x+1)$

If $( 2x-3)=0$

$\Rightarrow x=\frac{3}{2}$

If $( 3x+1)=0$

$\Rightarrow x=-\frac{1}{3}$

Thus, $x=\frac{3}{2},\ -\frac{1}{3}$

Sum of the roots$=\frac{3}{2}+( -\frac{1}{3})=\frac{9-2}{6}=\frac{7}{6}=-\frac{-b}{a}$   [Verified]

Product of the roots$=\frac{3}{2}\times( -\frac{1}{3})=-\frac{3}{6}=-\frac{1}{2}=\frac{c}{a}$    [Verified]

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Updated on: 10-Oct-2022

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