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Find the zeros of the quadratic polynomial 7y^2 - \frac{11}{3}y -\frac{2}{3} and verify the relationship between the zeroes and the coefficients.

 

 

Answers (1)

Given quadratic polynomial

    \begin{align*} p(y) &= 7y^2 -\frac{11}{3}y - \frac{2}{3}\\& = \frac{1}{3}\left[(21y^2 - 11y - 2) \right ] \\ & = \frac{1}{3}\left[(7y+1)(3y -2) \right ] \end{align*}

So, zeroes are

 \frac{2}{3}, \frac{-1}{7}

\Rightarrow Sum of zeros

  = \frac{2}{3} - \frac{1}{7} = \frac{11}{21}

   \frac{-b}{a} = \frac{11}{21}    \therefore Sum of zeros =\frac{-b}{a}

\Rightarrow Product of zeros

 = \left(\frac{2}{3} \right )\left(\frac{-1}{7} \right ) = \frac{-2}{21}

        \frac{c}{a} = \frac{-2}{21}\quad \thereforeproduct =\frac{c}{a}

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Safeer PP

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