If the equation $x^{2} - (2 + m) x + (- m^{2} - 4 m - 4) = 0$ has coincident roots, then

m = 0, m = 1

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m = 2, m = 2

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m = - 2, m = - 2

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m = 6, m = 1

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Solution

The correct option is D $m = - 2, m = - 2$
We know that for $a x^{2} + b x + c = 0$ , if the roots are equal, then

$b^{2} - 4 a c = 0$

Consider the equation given in the question.

$x^{2} - (2 + m) x + (- m^{2} - 4 m - 4) = 0$

Here,

$a = 1$

$b = - (2 + m)$

$c = (- m^{2} - 4 m - 4)$

Therefore,

${(2 + m)}^{2} - 4 \times (- m^{2} - 4 m - 4) = 0$

$4 + m^{2} + 4 m + 4 m^{2} + 16 m + 16 = 0$

$5 m^{2} + 20 m + 20 = 0$

$m^{2} + 4 m + 4 = 0$

${(m + 2)}^{2} = 0$

$m = - 2$
Hence, this is the required result.

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