Question

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

$7y^2-\frac{11}{3}y-\frac{2}{3}$

• A. $\frac{1}{3},\frac{-1}{7}$
• B. $\frac{2}{3},\frac{-1}{7}$
• C. $\frac{2}{3},\frac{-2}{7}$
• D. $\frac{2}{3},\frac{2}{7}$

Answer 1

Answer: (B)

$\frac{2}{3},\frac{-1}{7}$

Given quadratic polynomial is $7y^2-\frac{11}{3}y-\frac{2}{3}$.

$=\frac{1}{3}(21y^2-11y-2)$

$=\frac{1}{3}(21y^2-14y+3y-2)$

$=\frac{1}{3}\left[7y\right(3y-2)+(3y-2\left)\right]$

$=\frac{1}{3}(3y-2)(7y+1)$

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

The zeroes of the polynomials are,

$\frac{2}{3}$ , $-\frac{1}{7}$

Relationship between the zeroes and the coefficients of the polynomials-

Sum of the zeros=-$\frac{coefficientofy}{coefficientof{y}^2}=-\left(\frac{-\frac{11}{3}}{7}\right)=\frac{11}{21}$

Also sum of zeroes= $\frac{2}{3}+(-\frac{1}{7})$

=$\frac{14-3}{21}$

=$\frac{11}{21}$

Product of the zeroes =$\frac{constantterm}{coefficientof{y}^2}=\frac{-\frac{2}{3}}{7}=\frac{-2}{21}$

Also the product of the zeroes=$\frac{2}{3}\times (-\frac{1}{7})$=$\frac{-2}{21}$

Hence verified.

$OptionBiscorrect$