Roots of quadratic equation $3 y^{2} + 7 y + 4 = 0$ are

y = - 1, \frac{- 4}{3}

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y = 2, \frac{- 4}{3}

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y = - 2, \frac{- 4}{3}

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y = - 1, \frac{- 4}{5}

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Solution

The correct option is D $y = - 1, \frac{- 4}{3}$
Given equation is
$3 y^{2} + 7 y + 4 = 0$
$3 y^{2} + 3 y + 4 y + 4 = 0$
$3 y (y + 1) + 4 (y + 1) = 0$
$(3 y + 4) (y + 1) = 0$
$∴ y = \frac{- 4}{3}, - 1$
$∴$ the roots of the given equation are $\frac{- 4}{3}, - 1$

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