Solve : $2 x^{2} + 7 x + 3 = 0$

x = - \frac{1}{2}, 3

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x = - \frac{1}{2}, - 3

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x = \frac{1}{2}, - 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = \frac{1}{2}, 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

The correct option is A $x = - \frac{1}{2}, - 3$
$2 x^{2} + 7 x + 3 = 0 \Rightarrow 2 x^{2} + 6 x + x + 3 = 0 \Rightarrow 2 x (x + 3) + 1 (x + 3) = 0 \Rightarrow (x + 3) (2 x + 1) = 0$
$\Rightarrow x = - 3$ or $x = \frac{- 1}{2}$

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\frac{4 x - 3}{2 x + 1} - 10 (\frac{2 x + 1}{4 x - 3}) = 3; x \neq - \frac{1}{2}, \frac{3}{4}

\frac{4 x - 3}{2 x + 1} - 10 (\frac{2 x + 1}{4 x - 3}) = 3, x \neq \frac{- 1}{2}, \frac{3}{4}

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3 x^{2} + 2 x - 1 = 0

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