Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and their coefficients:
$q(y)\ =\ 7y^{2} \ –\ \left(\frac{11}{3}\right) y\ –\ \frac{2}{3}$

Given:


$q(y) = 7y^2-(\frac{11}{3})y-\frac{2}{3}$

To find:

Here, we have to find the zeros of q(y). 

Solution:

To find the zeros of q(y), we have to put $q(y)=0$.

This implies,

$7y^2-(\frac{11}{3})y-\frac{2}{3}= 0$

Multiplying by 3 on both sides, 

$3(7y^2)-3(\frac{11}{3})y-3(\frac{2}{3})= 0$

$21y^2-11y-2= 0$

$21y^2-14y+3y-2= 0$

$7y(3y-2) +1(3y-2) =0$

$(7y+1) (3y-2) =0$

$7y+1=0$ and $3y-2=0$

$7y= -1$ and $3y= 2$

$y=-\frac{1}{7}$ and $y=\frac{2}{3}$

Therefore, the zeros of the quadratic equation $q(y) = 7y^2-(\frac{11}{3})y-\frac{2}{3}$ are $-\frac{1}{7}$ and $\frac{2}{3}$.

Verification:

We know that, 
Sum of zeros $= -\frac{coefficient of y}{coefficient of y^2}$

                       $= –(\frac{\frac{-11}{3}}{7})$

                       $=\frac{11}{21}$

Sum of the zeros of $q(y)=-\frac{1}{7}+\frac{2}{3}=\frac{-1\times3+2\times7}{21}=\frac{-3+14}{21}=\frac{11}{21}$ 

Product of roots $= \frac{constant}{coefficient of y^2}$

                            $=\frac{\frac{-2}{3}}{7}$

                            $= -\frac{2}{21}$

Product of the roots of $q(y)=-\frac{1}{7}\times \frac{2}{3} =-\frac{2}{21}$

Hence, the relationship between the zeros and their coefficients is verified.

Updated on: 10-Oct-2022

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