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Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and their coefficients:

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

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\frac{1}{- 7}, \frac{3}{4}

\frac{3}{4}, \frac{5}{3}

\frac{1}{7}, \frac{- 2}{5}

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

answer is D.

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Detailed Solution

Given, quadratic function,

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

We put the value

q (y) = 0

to get,

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

\Rightarrow \frac{(21 y^{2} - 11 y - 2)}{3} = 0

\Rightarrow 21 y^{2} - 11 y - 2 = 0

\Rightarrow 21 y^{2} - 14 y + 3 y - 2 = 0

\Rightarrow 7 y (3 y - 2) - 1 (3 y + 2) = 0

\Rightarrow (3 y - 2) (7 y + 1) = 0

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

Thus, zeros of the polynomial

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

For the verification of the relation between 0 and their coefficients
Using this formula,

Sum of zeros = - \frac{coefficient of y}{coefficient of y^{2}}

\frac{2}{3} + (- \frac{1}{7}) = - \frac{- \frac{11}{3}}{7}

- \frac{11}{21} = - \frac{11}{21}

We calculate

Product of roots

= \frac{constant}{coefficient of y^{2}}

\frac{2}{3} \times (- \frac{1}{7}) = \frac{- \frac{2}{3}}{7}

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

Verified the relation between zeros and their coefficients.
Hence, the zeros of each of

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

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

.
Option 4 is correct.

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