If $l o g_{10} (2 x^{2} + 7 x + 16) = 1$ , then value of x is

-2 or

\frac{3}{2}

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-2 or

- \frac{3}{2}

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-2 or

\frac{2}{3}

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2 or

\frac{3}{2}

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Solution

The correct option is B -2 or $- \frac{3}{2}$
Consider the given equation,

${log}_{10} (2 x^{2} + 7 x + 16) = 1$

$2 x^{2} + 7 x + 16 = 10^{1}$

$2 x^{2} + 7 x + 16 = 10$

$2 x^{2} + 7 x + 6 = 0$

$2 x^{2} + 4 x + 3 x + 6 = 0$

$2 x (x + 2) + 3 (x + 2) = 0$

$(x + 2) (2 x + 3) = 0$

When,

$x + 2 = 0$

$x = - 2$

When,

$2 x + 3 = 0$

$x = \frac{- 3}{2}$

$x = - 2 o r \frac{- 3}{2}$

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