The value of m (m being real) for which the equation $m x^{2} + 2 x + m = 0$ has two distinct real roots is

m = -1

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m = 1

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m = 2

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m = 0

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Solution

The correct options are
A m = -1
B m = 1
For $m x^{2} + 2 x + m = 0$ ,
$D = b^{2} - 4 a c = (2)^{2} - 4 \times m \times m = 4 - 4 m^{2}$

The roots of quadratic equation are real and distinct only when $D = 0$ . $\Rightarrow 4 - 4 m^{2} = 0$
$\Rightarrow 4 (1 - m^{2}) = 0$
$\Rightarrow (1 + m) (1 - m) = 0$
$\Rightarrow (1 + m) = 0 or (1 - m) = 0$
$\Rightarrow m = - 1 or m = 1$

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