The minimum value of the quadratic expression $x^{2} + 2 b x + c$ is

c b^{2}

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c^{2} b

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c + b^{2}

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c - b^{2}

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Solution

The correct option is D $c - b^{2}$
$f (x) = x^{2} + 2 b x + c$
Differentiating the above function with respect to x.
$f^{'} (x) = 2 x + 2 b$
$= 0$
Or
$2 x + 2 b = 0$
$x = - b$ .
Now
$f^{''} (x) = 2$
or
$f^{''} (x)_{x = - b} > 0$
Hence the function attains a minimum value at $x$ equal to $- b$
Therefore
$f (- b) = b^{2} - 2 b^{2} + c$
$= c - b^{2}$

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